3x^2+8x-435=0

Solution for 3x^2+8x-435=0 equation:

3x^2+8x-435=0

a = 3; b = 8; c = -435;
Δ = b2-4ac
Δ = 82-4·3·(-435)
Δ = 5284
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:

$\sqrt{\Delta}=\sqrt{5284}=\sqrt{4*1321}=\sqrt{4}*\sqrt{1321}=2\sqrt{1321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{1321}}{2*3}=\frac{-8-2\sqrt{1321}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{1321}}{2*3}=\frac{-8+2\sqrt{1321}}{6}
$