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2h^2+5=12
We move all terms to the left:
2h^2+5-(12)=0
We add all the numbers together, and all the variables
2h^2-7=0
a = 2; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·2·(-7)
Δ = 56
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*2}=\frac{0-2\sqrt{14}}{4} =-\frac{2\sqrt{14}}{4} =-\frac{\sqrt{14}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*2}=\frac{0+2\sqrt{14}}{4} =\frac{2\sqrt{14}}{4} =\frac{\sqrt{14}}{2} $
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