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There are different ways to solve a quadratic equation.

If you can see how to factorise the quadratic expression then this should be the quickest method.

For this example we have

\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]

So either \\[\\simplify{ ({a} * x + { -c}) =0} \\ \\text{or} \\ \\ \\simplify{ ({b} * x + { -d})}=0\\]

So the roots are:
\\[x= \\simplify{{n1-n4}/{2*a*b}} \\ \\ \\text{and} \\ \\ x= \\simplify{{n1+n4}/{2*a*b}} \\]

If you cannot see a factorisation then you can use the quadratic formula:

\\[ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\\]

Remember that there are three possible types of solution, depending upon the value of the discriminant $\\Delta=b^2-4ac$.

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. (The roots are complex)

For this question the discriminant is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$. {rdis}

So the {rep} roots are:

\\[ x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} = \\frac{\\var{n1} - \\var{n4} }{\\var{n3}} = \\simplify{{n1 - n4}/ {n3}} \\ \\ \\text{and} \\ \\ x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} = \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} = \\simplify{{n1 + n4}/ {n3}} \\]

You could also use the method of completing the square.

First we write:
\\[\\begin{eqnarray} \\simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\\var{n5}\\left(\\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2+ \\simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2 -\\simplify{ {n2^2}/{4*(a*b)^2}}\\right) \\end{eqnarray} \\]
So we need to solve:
\\[\\begin{eqnarray} \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}& -\\simplify{ {n2^2}/{4*(a*b)^2}}=0\\Rightarrow\\\\ \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}&=\\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \\end{eqnarray}\\]
and get the two {rep} solutions:
\\[\\begin{eqnarray} \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{-{abs(n2)}/{2*a*b}} \\Rightarrow &x& = \\simplify{({-abs(n2)+n1}/{2*a*b})}\\\\ \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{({abs(n2)}/{2*a*b})} \\Rightarrow &x& = \\simplify{({n1+abs(n2)}/{2*a*b})} \\end{eqnarray}\\]

See also these Mathcentre leaflets on solving quadratic equations.

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Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

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Znajd\u017a pierwiastki podanego równania kwadratowego

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\\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]

Podja warto\u015bci pierwiastków jako u\u0142amki lub liczby ca\u0142kowite, nie dziesi\u0119tne

Wpisz pierwiastek z ni\u017csz\u0105 warto\u015bci\u0105 jako pierwszy, je\u017celi równanie posiada jeden podwójny pierwiastek wpisz go w obu miejscach.

Pierwiuastek z ni\u017csz\u0105 warto\u015bci\u0105 $x=\\;$ [[0]].

Pierwiastek z wy\u017csza warto\u015bci\u0105 $x=\\;$ [[1]]

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Input numbers as fractions or integers not as a decimals.

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