6x(x+3)+10=9x-3(x-5)

Solution for 6x(x+3)+10=9x-3(x-5) equation:

6x(x+3)+10=9x-3(x-5)

We move all terms to the left:

6x(x+3)+10-(9x-3(x-5))=0

We multiply parentheses

6x^2+18x-(9x-3(x-5))+10=0


We calculate terms in parentheses: -(9x-3(x-5)), so:

9x-3(x-5)

We multiply parentheses

9x-3x+15

We add all the numbers together, and all the variables

6x+15

Back to the equation:

-(6x+15)


We get rid of parentheses

6x^2+18x-6x-15+10=0

We add all the numbers together, and all the variables

6x^2+12x-5=0

a = 6; b = 12; c = -5;
Δ = b2-4ac
Δ = 122-4·6·(-5)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{66}}{2*6}=\frac{-12-2\sqrt{66}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{66}}{2*6}=\frac{-12+2\sqrt{66}}{12}
$