3x^2+24x+7=23

Solution for 3x^2+24x+7=23 equation:

3x^2+24x+7=23

We move all terms to the left:

3x^2+24x+7-(23)=0

We add all the numbers together, and all the variables

3x^2+24x-16=0

a = 3; b = 24; c = -16;
Δ = b2-4ac
Δ = 242-4·3·(-16)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{3}}{2*3}=\frac{-24-16\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{3}}{2*3}=\frac{-24+16\sqrt{3}}{6}
$