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2x^2-4x=28
We move all terms to the left:
2x^2-4x-(28)=0
a = 2; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·2·(-28)
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{15}}{2*2}=\frac{4-4\sqrt{15}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{15}}{2*2}=\frac{4+4\sqrt{15}}{4} $
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