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(H)=3H^2-6H-15
We move all terms to the left:
(H)-(3H^2-6H-15)=0
We get rid of parentheses
-3H^2+H+6H+15=0
We add all the numbers together, and all the variables
-3H^2+7H+15=0
a = -3; b = 7; c = +15;
Δ = b2-4ac
Δ = 72-4·(-3)·15
Δ = 229
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{229}}{2*-3}=\frac{-7-\sqrt{229}}{-6} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{229}}{2*-3}=\frac{-7+\sqrt{229}}{-6} $
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