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4m^2+8m-12=0.
a = 4; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·4·(-12)
Δ = 256
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{256}=16$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*4}=\frac{-24}{8} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*4}=\frac{8}{8} =1 $
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