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3h^2-142=0
a = 3; b = 0; c = -142;
Δ = b2-4ac
Δ = 02-4·3·(-142)
Δ = 1704
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1704}=\sqrt{4*426}=\sqrt{4}*\sqrt{426}=2\sqrt{426}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{426}}{2*3}=\frac{0-2\sqrt{426}}{6} =-\frac{2\sqrt{426}}{6} =-\frac{\sqrt{426}}{3} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{426}}{2*3}=\frac{0+2\sqrt{426}}{6} =\frac{2\sqrt{426}}{6} =\frac{\sqrt{426}}{3} $
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