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=152-16H^2
We move all terms to the left:
-(152-16H^2)=0
We get rid of parentheses
16H^2-152=0
a = 16; b = 0; c = -152;
Δ = b2-4ac
Δ = 02-4·16·(-152)
Δ = 9728
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9728}=\sqrt{256*38}=\sqrt{256}*\sqrt{38}=16\sqrt{38}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{38}}{2*16}=\frac{0-16\sqrt{38}}{32} =-\frac{16\sqrt{38}}{32} =-\frac{\sqrt{38}}{2} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{38}}{2*16}=\frac{0+16\sqrt{38}}{32} =\frac{16\sqrt{38}}{32} =\frac{\sqrt{38}}{2} $
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