(2x+5)(3x+15)=(x+10)

Solution for (2x+5)(3x+15)=(x+10) equation:

(2x+5)(3x+15)=(x+10)

We move all terms to the left:

(2x+5)(3x+15)-((x+10))=0

We multiply parentheses ..

(+6x^2+30x+15x+75)-((x+10))=0


We calculate terms in parentheses: -((x+10)), so:

(x+10)

We get rid of parentheses

x+10

Back to the equation:

-(x+10)


We get rid of parentheses

6x^2+30x+15x-x+75-10=0

We add all the numbers together, and all the variables

6x^2+44x+65=0

a = 6; b = 44; c = +65;
Δ = b2-4ac
Δ = 442-4·6·65
Δ = 376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-2\sqrt{94}}{2*6}=\frac{-44-2\sqrt{94}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+2\sqrt{94}}{2*6}=\frac{-44+2\sqrt{94}}{12}
$