(m+3)(m-4)+10=0

Solution for (m+3)(m-4)+10=0 equation:

(m+3)(m-4)+10=0

We multiply parentheses ..

(+m^2-4m+3m-12)+10=0

We get rid of parentheses

m^2-4m+3m-12+10=0

We add all the numbers together, and all the variables

m^2-1m-2=0

a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2}
=-1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2}
=2
$