4x^2+28x-45=0

Solution for 4x^2+28x-45=0 equation:

4x^2+28x-45=0

a = 4; b = 28; c = -45;
Δ = b2-4ac
Δ = 282-4·4·(-45)
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{94}}{2*4}=\frac{-28-4\sqrt{94}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{94}}{2*4}=\frac{-28+4\sqrt{94}}{8}
$