2(m+3)+5(m2)=23

Solution for 2(m+3)+5(m2)=23 equation:

2(m+3)+5(m2)=23

We move all terms to the left:

2(m+3)+5(m2)-(23)=0

We add all the numbers together, and all the variables

5m^2+2(m+3)-23=0

We multiply parentheses

5m^2+2m+6-23=0

We add all the numbers together, and all the variables

5m^2+2m-17=0

a = 5; b = 2; c = -17;
Δ = b2-4ac
Δ = 22-4·5·(-17)
Δ = 344
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:

$\sqrt{\Delta}=\sqrt{344}=\sqrt{4*86}=\sqrt{4}*\sqrt{86}=2\sqrt{86}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{86}}{2*5}=\frac{-2-2\sqrt{86}}{10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{86}}{2*5}=\frac{-2+2\sqrt{86}}{10}
$