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2m^2+7m-13=-10
We move all terms to the left:
2m^2+7m-13-(-10)=0
We add all the numbers together, and all the variables
2m^2+7m-3=0
a = 2; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·2·(-3)
Δ = 73
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{-(7)-\sqrt{73}}{2*2}=\frac{-7-\sqrt{73}}{4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{73}}{2*2}=\frac{-7+\sqrt{73}}{4} $
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