2(3x2)=2x+28

Solution for 2(3x2)=2x+28 equation:

2(3x^2)=2x+28

We move all terms to the left:

2(3x^2)-(2x+28)=0

We get rid of parentheses

23x^2-2x-28=0

a = 23; b = -2; c = -28;
Δ = b2-4ac
Δ = -22-4·23·(-28)
Δ = 2580
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:

$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{645}}{2*23}=\frac{2-2\sqrt{645}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{645}}{2*23}=\frac{2+2\sqrt{645}}{46}
$