4x^2+28x-81=0

Solution for 4x^2+28x-81=0 equation:

4x^2+28x-81=0

a = 4; b = 28; c = -81;
Δ = b2-4ac
Δ = 282-4·4·(-81)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{130}}{2*4}=\frac{-28-4\sqrt{130}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{130}}{2*4}=\frac{-28+4\sqrt{130}}{8}
$