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2x^2-16=10x
We move all terms to the left:
2x^2-16-(10x)=0
a = 2; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·2·(-16)
Δ = 228
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{57}}{2*2}=\frac{10-2\sqrt{57}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{57}}{2*2}=\frac{10+2\sqrt{57}}{4} $
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