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h(h+10)-28=0
We multiply parentheses
h^2+10h-28=0
a = 1; b = 10; c = -28;
Δ = b2-4ac
Δ = 102-4·1·(-28)
Δ = 212
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{53}}{2*1}=\frac{-10-2\sqrt{53}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{53}}{2*1}=\frac{-10+2\sqrt{53}}{2} $
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