3x^2+42x-150=0

Solution for 3x^2+42x-150=0 equation:

3x^2+42x-150=0

a = 3; b = 42; c = -150;
Δ = b2-4ac
Δ = 422-4·3·(-150)
Δ = 3564
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:

$\sqrt{\Delta}=\sqrt{3564}=\sqrt{324*11}=\sqrt{324}*\sqrt{11}=18\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-18\sqrt{11}}{2*3}=\frac{-42-18\sqrt{11}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+18\sqrt{11}}{2*3}=\frac{-42+18\sqrt{11}}{6}
$