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=4+60H-16H^2
We move all terms to the left:
-(4+60H-16H^2)=0
We get rid of parentheses
16H^2-60H-4=0
a = 16; b = -60; c = -4;
Δ = b2-4ac
Δ = -602-4·16·(-4)
Δ = 3856
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3856}=\sqrt{16*241}=\sqrt{16}*\sqrt{241}=4\sqrt{241}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{241}}{2*16}=\frac{60-4\sqrt{241}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{241}}{2*16}=\frac{60+4\sqrt{241}}{32} $
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