4(3x+5)=5(x+4)7x

Solution for 4(3x+5)=5(x+4)7x equation:

4(3x+5)=5(x+4)7x

We move all terms to the left:

4(3x+5)-(5(x+4)7x)=0

We multiply parentheses

12x-(5(x+4)7x)+20=0


We calculate terms in parentheses: -(5(x+4)7x), so:

5(x+4)7x

We multiply parentheses

35x^2+140x

Back to the equation:

-(35x^2+140x)


We get rid of parentheses

-35x^2+12x-140x+20=0

We add all the numbers together, and all the variables

-35x^2-128x+20=0

a = -35; b = -128; c = +20;
Δ = b2-4ac
Δ = -1282-4·(-35)·20
Δ = 19184
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:

$\sqrt{\Delta}=\sqrt{19184}=\sqrt{16*1199}=\sqrt{16}*\sqrt{1199}=4\sqrt{1199}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-4\sqrt{1199}}{2*-35}=\frac{128-4\sqrt{1199}}{-70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+4\sqrt{1199}}{2*-35}=\frac{128+4\sqrt{1199}}{-70}
$