How do you solve $x^2 +5x-7=0$ ?

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Answer 1 · 2016-05-24T23:10:44

$x = (-5+-sqrt(53))/2$

This quadratic equation is in the form $ax^2+bx+c = 0$ with $a=1$ , $b=5$ and $c=-7$ .

It has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 5^2-(4*1*-7) = 25+28 = 53$

So $Delta > 0$ , but it is not a perfect square, so our quadratic equation has irrational roots.

We can find them using the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-b+-sqrt(Delta))/(2a)$

$=(-5+-sqrt(53))/2$

How do you solve x^2 +5x-7=0x^2 +5x-7=0?