x^2+3x-4=28

Solution for x^2+3x-4=28 equation:

x^2+3x-4=28

We move all terms to the left:

x^2+3x-4-(28)=0

We add all the numbers together, and all the variables

x^2+3x-32=0

a = 1; b = 3; c = -32;
Δ = b2-4ac
Δ = 32-4·1·(-32)
Δ = 137
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{-(3)-\sqrt{137}}{2*1}=\frac{-3-\sqrt{137}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{137}}{2*1}=\frac{-3+\sqrt{137}}{2}
$