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