1m^2+17m+5=0

Solution for 1m^2+17m+5=0 equation:

1m^2+17m+5=0

We add all the numbers together, and all the variables

m^2+17m+5=0

a = 1; b = 17; c = +5;
Δ = b2-4ac
Δ = 172-4·1·5
Δ = 269
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{-(17)-\sqrt{269}}{2*1}=\frac{-17-\sqrt{269}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{269}}{2*1}=\frac{-17+\sqrt{269}}{2}
$