(x+3)(2x+5)=10

Solution for (x+3)(2x+5)=10 equation:

(x+3)(2x+5)=10

We move all terms to the left:

(x+3)(2x+5)-(10)=0

We multiply parentheses ..

(+2x^2+5x+6x+15)-10=0

We get rid of parentheses

2x^2+5x+6x+15-10=0

We add all the numbers together, and all the variables

2x^2+11x+5=0

a = 2; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·2·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*2}=\frac{-2}{4}
=-1/2
$