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=-16H^2-16H+160
We move all terms to the left:
-(-16H^2-16H+160)=0
We get rid of parentheses
16H^2+16H-160=0
a = 16; b = 16; c = -160;
Δ = b2-4ac
Δ = 162-4·16·(-160)
Δ = 10496
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{10496}=\sqrt{256*41}=\sqrt{256}*\sqrt{41}=16\sqrt{41}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{41}}{2*16}=\frac{-16-16\sqrt{41}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{41}}{2*16}=\frac{-16+16\sqrt{41}}{32} $
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