2x(3-x)+1=-5(2x+3)+3

Solution for 2x(3-x)+1=-5(2x+3)+3 equation:

2x(3-x)+1=-5(2x+3)+3

We move all terms to the left:

2x(3-x)+1-(-5(2x+3)+3)=0

We add all the numbers together, and all the variables

2x(-1x+3)-(-5(2x+3)+3)+1=0

We multiply parentheses

-2x^2+6x-(-5(2x+3)+3)+1=0


We calculate terms in parentheses: -(-5(2x+3)+3), so:

-5(2x+3)+3

We multiply parentheses

-10x-15+3

We add all the numbers together, and all the variables

-10x-12

Back to the equation:

-(-10x-12)


We get rid of parentheses

-2x^2+6x+10x+12+1=0

We add all the numbers together, and all the variables

-2x^2+16x+13=0

a = -2; b = 16; c = +13;
Δ = b2-4ac
Δ = 162-4·(-2)·13
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{10}}{2*-2}=\frac{-16-6\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{10}}{2*-2}=\frac{-16+6\sqrt{10}}{-4}
$