m^2+7m-297=0

Solution for m^2+7m-297=0 equation:

m^2+7m-297=0

a = 1; b = 7; c = -297;
Δ = b2-4ac
Δ = 72-4·1·(-297)
Δ = 1237
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{1237}}{2*1}=\frac{-7-\sqrt{1237}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{1237}}{2*1}=\frac{-7+\sqrt{1237}}{2}
$