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=-15H^2+135H+10
We move all terms to the left:
-(-15H^2+135H+10)=0
We get rid of parentheses
15H^2-135H-10=0
a = 15; b = -135; c = -10;
Δ = b2-4ac
Δ = -1352-4·15·(-10)
Δ = 18825
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18825}=\sqrt{25*753}=\sqrt{25}*\sqrt{753}=5\sqrt{753}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-135)-5\sqrt{753}}{2*15}=\frac{135-5\sqrt{753}}{30} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-135)+5\sqrt{753}}{2*15}=\frac{135+5\sqrt{753}}{30} $
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