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

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

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

We move all terms to the left:

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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2-1x+2=0

a = -4; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-4)·2
Δ = 33
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{-(-1)-\sqrt{33}}{2*-4}=\frac{1-\sqrt{33}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{33}}{2*-4}=\frac{1+\sqrt{33}}{-8}
$