4x^2+36x-28=0

Solution for 4x^2+36x-28=0 equation:

4x^2+36x-28=0

a = 4; b = 36; c = -28;
Δ = b2-4ac
Δ = 362-4·4·(-28)
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{109}}{2*4}=\frac{-36-4\sqrt{109}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{109}}{2*4}=\frac{-36+4\sqrt{109}}{8}
$