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4m^2+8m+3=0
a = 4; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·4·3
Δ = 16
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*4}=\frac{-12}{8} =-1+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*4}=\frac{-4}{8} =-1/2 $
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