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3a^2+8a=4
We move all terms to the left:
3a^2+8a-(4)=0
a = 3; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·3·(-4)
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{7}}{2*3}=\frac{-8-4\sqrt{7}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{7}}{2*3}=\frac{-8+4\sqrt{7}}{6} $
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