How do you solve $x^{2} - 7 x + 12 = 0$ $x^2- 7x + 12 = 0$ using the quadratic formula?

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How do you solve $x^{2} - 7 x + 12 = 0$ $x^2- 7x + 12 = 0$ using the quadratic formula?

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Answer 1 · 2018-05-02T16:20:43

$x = 4$ $x=4$ or $3$ $3$

Explanation:

According to quadratic formula if $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , then $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Here we have $x^{2} - 7 x + 12 = 0$ $x^2-7x+12=0$ i.e. $a = 1$ $a=1$ , $b = - 7$ $b=-7$ and $c = 12$ $c=12$

and $x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$ $x=(-(-7)+-sqrt((-7)^2-4*1*12))/(2*1)$

= $\frac{7 \pm \sqrt{49 - 48}}{2}$ $(7+-sqrt(49-48))/2$

= $\frac{7 \pm \sqrt{1}}{2}$ $(7+-sqrt1)/2$

= $\frac{7 \pm 1}{2}$ $(7+-1)/2$

i.e. $x = 4$ $x=4$ or $3$ $3$

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How do you solve $x^{2} - 7 x + 12 = 0$ $x^2- 7x + 12 = 0$ using the quadratic formula?

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Explanation:

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