-5(1x+2)=2(2x-15)1x

Solution for -5(1x+2)=2(2x-15)1x equation:

-5(1x+2)=2(2x-15)1x

We move all terms to the left:

-5(1x+2)-(2(2x-15)1x)=0

We add all the numbers together, and all the variables

-5(x+2)-(2(2x-15)1x)=0

We multiply parentheses

-5x-(2(2x-15)1x)-10=0


We calculate terms in parentheses: -(2(2x-15)1x), so:

2(2x-15)1x

We multiply parentheses

4x^2-30x

Back to the equation:

-(4x^2-30x)


We get rid of parentheses

-4x^2-5x+30x-10=0

We add all the numbers together, and all the variables

-4x^2+25x-10=0

a = -4; b = 25; c = -10;
Δ = b2-4ac
Δ = 252-4·(-4)·(-10)
Δ = 465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{465}}{2*-4}=\frac{-25-\sqrt{465}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{465}}{2*-4}=\frac{-25+\sqrt{465}}{-8}
$