x,5(2x+3)=3(3x+5)

Solution for x,5(2x+3)=3(3x+5) equation:

x.5(2x+3)=3(3x+5)

We move all terms to the left:

x.5(2x+3)-(3(3x+5))=0

We multiply parentheses

2x^2+3x-(3(3x+5))=0


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

3(3x+5)

We multiply parentheses

9x+15

Back to the equation:

-(9x+15)


We get rid of parentheses

2x^2+3x-9x-15=0

We add all the numbers together, and all the variables

2x^2-6x-15=0

a = 2; b = -6; c = -15;
Δ = b2-4ac
Δ = -62-4·2·(-15)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{39}}{2*2}=\frac{6-2\sqrt{39}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{39}}{2*2}=\frac{6+2\sqrt{39}}{4}
$