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(H)=-2H^2-7H+4
We move all terms to the left:
(H)-(-2H^2-7H+4)=0
We get rid of parentheses
2H^2+7H+H-4=0
We add all the numbers together, and all the variables
2H^2+8H-4=0
a = 2; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·2·(-4)
Δ = 96
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{6}}{2*2}=\frac{-8-4\sqrt{6}}{4} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{6}}{2*2}=\frac{-8+4\sqrt{6}}{4} $
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