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2x(2x+2)=312
We move all terms to the left:
2x(2x+2)-(312)=0
We multiply parentheses
4x^2+4x-312=0
a = 4; b = 4; c = -312;
Δ = b2-4ac
Δ = 42-4·4·(-312)
Δ = 5008
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5008}=\sqrt{16*313}=\sqrt{16}*\sqrt{313}=4\sqrt{313}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{313}}{2*4}=\frac{-4-4\sqrt{313}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{313}}{2*4}=\frac{-4+4\sqrt{313}}{8} $
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