-3x^2+8x+40=23

Solution for -3x^2+8x+40=23 equation:

-3x^2+8x+40=23

We move all terms to the left:

-3x^2+8x+40-(23)=0

We add all the numbers together, and all the variables

-3x^2+8x+17=0

a = -3; b = 8; c = +17;
Δ = b2-4ac
Δ = 82-4·(-3)·17
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{67}}{2*-3}=\frac{-8-2\sqrt{67}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{67}}{2*-3}=\frac{-8+2\sqrt{67}}{-6}
$