3x^2+10x-150=0

Solution for 3x^2+10x-150=0 equation:

3x^2+10x-150=0

a = 3; b = 10; c = -150;
Δ = b2-4ac
Δ = 102-4·3·(-150)
Δ = 1900
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:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{19}}{2*3}=\frac{-10-10\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{19}}{2*3}=\frac{-10+10\sqrt{19}}{6}
$