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h^2+21=-10h
We move all terms to the left:
h^2+21-(-10h)=0
We get rid of parentheses
h^2+10h+21=0
a = 1; b = 10; c = +21;
Δ = b2-4ac
Δ = 102-4·1·21
Δ = 16
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*1}=\frac{-14}{2} =-7 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*1}=\frac{-6}{2} =-3 $
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