m^2+7m-60=0

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Solution for m^2+7m-60=0 equation: 



m^2+7m-60=0

a = 1; b = 7; c = -60;
Δ = b2-4ac
Δ = 72-4·1·(-60)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*1}=\frac{-24}{2}
=-12
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*1}=\frac{10}{2}
=5
$

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