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2m^2+15m-15=0
a = 2; b = 15; c = -15;
Δ = b2-4ac
Δ = 152-4·2·(-15)
Δ = 345
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{345}}{2*2}=\frac{-15-\sqrt{345}}{4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{345}}{2*2}=\frac{-15+\sqrt{345}}{4} $
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