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8x^2+4x=10
We move all terms to the left:
8x^2+4x-(10)=0
a = 8; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·8·(-10)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{21}}{2*8}=\frac{-4-4\sqrt{21}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{21}}{2*8}=\frac{-4+4\sqrt{21}}{16} $
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