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4m^2+12m-9=0
a = 4; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·4·(-9)
Δ = 288
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{2}}{2*4}=\frac{-12-12\sqrt{2}}{8} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{2}}{2*4}=\frac{-12+12\sqrt{2}}{8} $
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