23x^2+33x-270=0

Solution for 23x^2+33x-270=0 equation:

23x^2+33x-270=0

a = 23; b = 33; c = -270;
Δ = b2-4ac
Δ = 332-4·23·(-270)
Δ = 25929
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:

$\sqrt{\Delta}=\sqrt{25929}=\sqrt{9*2881}=\sqrt{9}*\sqrt{2881}=3\sqrt{2881}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{2881}}{2*23}=\frac{-33-3\sqrt{2881}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{2881}}{2*23}=\frac{-33+3\sqrt{2881}}{46}
$