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=180-16H^2
We move all terms to the left:
-(180-16H^2)=0
We get rid of parentheses
16H^2-180=0
a = 16; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·16·(-180)
Δ = 11520
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{5}}{2*16}=\frac{0-48\sqrt{5}}{32} =-\frac{48\sqrt{5}}{32} =-\frac{3\sqrt{5}}{2} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{5}}{2*16}=\frac{0+48\sqrt{5}}{32} =\frac{48\sqrt{5}}{32} =\frac{3\sqrt{5}}{2} $
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