10x(x+1)-6x=9(x^2+5)

Solution for 10x(x+1)-6x=9(x^2+5) equation:

10x(x+1)-6x=9(x^2+5)

We move all terms to the left:

10x(x+1)-6x-(9(x^2+5))=0

We add all the numbers together, and all the variables

-6x+10x(x+1)-(9(x^2+5))=0

We multiply parentheses

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


We calculate terms in parentheses: -(9(x^2+5)), so:

9(x^2+5)

We multiply parentheses

9x^2+45

Back to the equation:

-(9x^2+45)


We add all the numbers together, and all the variables

10x^2+4x-(9x^2+45)=0

We get rid of parentheses

10x^2-9x^2+4x-45=0

We add all the numbers together, and all the variables

x^2+4x-45=0

a = 1; b = 4; c = -45;
Δ = b2-4ac
Δ = 42-4·1·(-45)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*1}=\frac{10}{2}
=5
$