24=(4x-5)(x+7)

Solution for 24=(4x-5)(x+7) equation:

24=(4x-5)(x+7)

We move all terms to the left:

24-((4x-5)(x+7))=0

We multiply parentheses ..

-((+4x^2+28x-5x-35))+24=0


We calculate terms in parentheses: -((+4x^2+28x-5x-35)), so:

(+4x^2+28x-5x-35)

We get rid of parentheses

4x^2+28x-5x-35

We add all the numbers together, and all the variables

4x^2+23x-35

Back to the equation:

-(4x^2+23x-35)


We get rid of parentheses

-4x^2-23x+35+24=0

We add all the numbers together, and all the variables

-4x^2-23x+59=0

a = -4; b = -23; c = +59;
Δ = b2-4ac
Δ = -232-4·(-4)·59
Δ = 1473
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{1473}}{2*-4}=\frac{23-\sqrt{1473}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1473}}{2*-4}=\frac{23+\sqrt{1473}}{-8}
$