25=(x+4)(x-3)

Solution for 25=(x+4)(x-3) equation:

25=(x+4)(x-3)

We move all terms to the left:

25-((x+4)(x-3))=0

We multiply parentheses ..

-((+x^2-3x+4x-12))+25=0


We calculate terms in parentheses: -((+x^2-3x+4x-12)), so:

(+x^2-3x+4x-12)

We get rid of parentheses

x^2-3x+4x-12

We add all the numbers together, and all the variables

x^2+x-12

Back to the equation:

-(x^2+x-12)


We get rid of parentheses

-x^2-x+12+25=0

We add all the numbers together, and all the variables

-1x^2-1x+37=0

a = -1; b = -1; c = +37;
Δ = b2-4ac
Δ = -12-4·(-1)·37
Δ = 149
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{-(-1)-\sqrt{149}}{2*-1}=\frac{1-\sqrt{149}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{149}}{2*-1}=\frac{1+\sqrt{149}}{-2}
$