3x^2+2x-20=3

Solution for 3x^2+2x-20=3 equation:

3x^2+2x-20=3

We move all terms to the left:

3x^2+2x-20-(3)=0

We add all the numbers together, and all the variables

3x^2+2x-23=0

a = 3; b = 2; c = -23;
Δ = b2-4ac
Δ = 22-4·3·(-23)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{70}}{2*3}=\frac{-2-2\sqrt{70}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{70}}{2*3}=\frac{-2+2\sqrt{70}}{6}
$