x^2+3x-4=25

Solution for x^2+3x-4=25 equation:

x^2+3x-4=25

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+3x-29=0

a = 1; b = 3; c = -29;
Δ = b2-4ac
Δ = 32-4·1·(-29)
Δ = 125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5\sqrt{5}}{2*1}=\frac{-3-5\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5\sqrt{5}}{2*1}=\frac{-3+5\sqrt{5}}{2}
$