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330=4h^2+10h
We move all terms to the left:
330-(4h^2+10h)=0
We get rid of parentheses
-4h^2-10h+330=0
a = -4; b = -10; c = +330;
Δ = b2-4ac
Δ = -102-4·(-4)·330
Δ = 5380
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5380}=\sqrt{4*1345}=\sqrt{4}*\sqrt{1345}=2\sqrt{1345}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{1345}}{2*-4}=\frac{10-2\sqrt{1345}}{-8} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{1345}}{2*-4}=\frac{10+2\sqrt{1345}}{-8} $
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