3-m*m+5m-7=m^2-3m-7

Solution for 3-m*m+5m-7=m^2-3m-7 equation:

3-m*m+5m-7=m^2-3m-7

We move all terms to the left:

3-m*m+5m-7-(m^2-3m-7)=0

We add all the numbers together, and all the variables

5m-m*m-(m^2-3m-7)-4=0

Wy multiply elements

-1m^2+5m-(m^2-3m-7)-4=0

We get rid of parentheses

-1m^2-m^2+5m+3m+7-4=0

We add all the numbers together, and all the variables

-2m^2+8m+3=0

a = -2; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·(-2)·3
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{22}}{2*-2}=\frac{-8-2\sqrt{22}}{-4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{22}}{2*-2}=\frac{-8+2\sqrt{22}}{-4}
$