4x^2+12x-435.25=0

Solution for 4x^2+12x-435.25=0 equation:

4x^2+12x-435.25=0

a = 4; b = 12; c = -435.25;
Δ = b2-4ac
Δ = 122-4·4·(-435.25)
Δ = 7108
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:

$\sqrt{\Delta}=\sqrt{7108}=\sqrt{4*1777}=\sqrt{4}*\sqrt{1777}=2\sqrt{1777}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{1777}}{2*4}=\frac{-12-2\sqrt{1777}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{1777}}{2*4}=\frac{-12+2\sqrt{1777}}{8}
$