-14+3(x+10)=7(2x+4)x+

Solution for -14+3(x+10)=7(2x+4)x+ equation:

-14+3(x+10)=7(2x+4)x+

We move all terms to the left:

-14+3(x+10)-(7(2x+4)x+)=0

We multiply parentheses

3x-(7(2x+4)x+)+30-14=0


We calculate terms in parentheses: -(7(2x+4)x+), so:

7(2x+4)x+

We add all the numbers together, and all the variables

7(2x+4)x

We multiply parentheses

14x^2+28x

Back to the equation:

-(14x^2+28x)


We add all the numbers together, and all the variables

3x-(14x^2+28x)+16=0

We get rid of parentheses

-14x^2+3x-28x+16=0

We add all the numbers together, and all the variables

-14x^2-25x+16=0

a = -14; b = -25; c = +16;
Δ = b2-4ac
Δ = -252-4·(-14)·16
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-39}{2*-14}=\frac{-14}{-28}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+39}{2*-14}=\frac{64}{-28}
=-2+2/7
$