How do you solve $x^{2} - 3 x - 18 = 0$ $x^2 - 3x - 18 = 0$ ?

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Answer 1 · 2015-11-02T21:19:09

$x_{1} = 6$ $x_1=6$ and $x_{2} = - 3$ $x_2=-3$

you have an Equation like this: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

Then you can use the formula for solving quadratic Equations:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

so we have: $x = - (- 3) (\pm) \frac{\sqrt{{(- 3)}^{2} - 4 \cdot 1 \cdot (- 18)}}{2 \cdot 1}$ $x=-(-3)(+-)sqrt((-3)^2-4*1*(-18))/(2*1)$

$x = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}$ $x=(3+-sqrt(9+72))/2=(3+-9)/2$

$x_{1} = \frac{3 + 9}{2} = 6$ $x_1=(3+9)/2=6$
$x_{2} = \frac{3 - 9}{2} = - 3$ $x_2=(3-9)/2=-3$

How do you solve x2−3x−18=0x^2 - 3x - 18 = 0?