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4m^2+15m+6=0
a = 4; b = 15; c = +6;
Δ = b2-4ac
Δ = 152-4·4·6
Δ = 129
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{129}}{2*4}=\frac{-15-\sqrt{129}}{8} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{129}}{2*4}=\frac{-15+\sqrt{129}}{8} $
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