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h^2+6h=15
We move all terms to the left:
h^2+6h-(15)=0
a = 1; b = 6; c = -15;
Δ = b2-4ac
Δ = 62-4·1·(-15)
Δ = 96
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{6}}{2*1}=\frac{-6-4\sqrt{6}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{6}}{2*1}=\frac{-6+4\sqrt{6}}{2} $
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