3x^2+40x-225=0

Solution for 3x^2+40x-225=0 equation:

3x^2+40x-225=0

a = 3; b = 40; c = -225;
Δ = b2-4ac
Δ = 402-4·3·(-225)
Δ = 4300
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:

$\sqrt{\Delta}=\sqrt{4300}=\sqrt{100*43}=\sqrt{100}*\sqrt{43}=10\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{43}}{2*3}=\frac{-40-10\sqrt{43}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{43}}{2*3}=\frac{-40+10\sqrt{43}}{6}
$