(3x-5)=(4-x^2)(3)

Solution for (3x-5)=(4-x^2)(3) equation:

(3x-5)=(4-x^2)(3)

We move all terms to the left:

(3x-5)-((4-x^2)(3))=0

We get rid of parentheses

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


We calculate terms in parentheses: -((4-x^2)3), so:

(4-x^2)3

We multiply parentheses

-3x^2+12

Back to the equation:

-(-3x^2+12)


We get rid of parentheses

3x^2+3x-12-5=0

We add all the numbers together, and all the variables

3x^2+3x-17=0

a = 3; b = 3; c = -17;
Δ = b2-4ac
Δ = 32-4·3·(-17)
Δ = 213
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{-(3)-\sqrt{213}}{2*3}=\frac{-3-\sqrt{213}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{213}}{2*3}=\frac{-3+\sqrt{213}}{6}
$