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12+15m-2m^2=0
a = -2; b = 15; c = +12;
Δ = b2-4ac
Δ = 152-4·(-2)·12
Δ = 321
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{321}}{2*-2}=\frac{-15-\sqrt{321}}{-4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{321}}{2*-2}=\frac{-15+\sqrt{321}}{-4} $
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