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3h^2+30h+63=0
a = 3; b = 30; c = +63;
Δ = b2-4ac
Δ = 302-4·3·63
Δ = 144
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{144}=12$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-12}{2*3}=\frac{-42}{6} =-7 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+12}{2*3}=\frac{-18}{6} =-3 $
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