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2x^2-32x=-60
We move all terms to the left:
2x^2-32x-(-60)=0
We add all the numbers together, and all the variables
2x^2-32x+60=0
a = 2; b = -32; c = +60;
Δ = b2-4ac
Δ = -322-4·2·60
Δ = 544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{34}}{2*2}=\frac{32-4\sqrt{34}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{34}}{2*2}=\frac{32+4\sqrt{34}}{4} $
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