(2x+10)x=(3x+16)(x+5)

Solution for (2x+10)x=(3x+16)(x+5) equation:

(2x+10)x=(3x+16)(x+5)

We move all terms to the left:

(2x+10)x-((3x+16)(x+5))=0

We multiply parentheses

2x^2+10x-((3x+16)(x+5))=0

We multiply parentheses ..

2x^2-((+3x^2+15x+16x+80))+10x=0


We calculate terms in parentheses: -((+3x^2+15x+16x+80)), so:

(+3x^2+15x+16x+80)

We get rid of parentheses

3x^2+15x+16x+80

We add all the numbers together, and all the variables

3x^2+31x+80

Back to the equation:

-(3x^2+31x+80)


We add all the numbers together, and all the variables

2x^2+10x-(3x^2+31x+80)=0

We get rid of parentheses

2x^2-3x^2+10x-31x-80=0

We add all the numbers together, and all the variables

-1x^2-21x-80=0

a = -1; b = -21; c = -80;
Δ = b2-4ac
Δ = -212-4·(-1)·(-80)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-11}{2*-1}=\frac{10}{-2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+11}{2*-1}=\frac{32}{-2}
=-16
$