(m^2+14m+49)-95=0

Solution for (m^2+14m+49)-95=0 equation:

(m^2+14m+49)-95=0

We get rid of parentheses

m^2+14m+49-95=0

We add all the numbers together, and all the variables

m^2+14m-46=0

a = 1; b = 14; c = -46;
Δ = b2-4ac
Δ = 142-4·1·(-46)
Δ = 380
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{95}}{2*1}=\frac{-14-2\sqrt{95}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{95}}{2*1}=\frac{-14+2\sqrt{95}}{2}
$