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

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Answer 1 · 2018-05-02T16:20:43

$x=4$ or $3$

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

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

and $x=(-(-7)+-sqrt((-7)^2-4*1*12))/(2*1)$

= $(7+-sqrt(49-48))/2$

= $(7+-sqrt1)/2$

= $(7+-1)/2$

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

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