4=3x^2+4x-5

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

4=3x^2+4x-5

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2-4x+9=0

a = -3; b = -4; c = +9;
Δ = b2-4ac
Δ = -42-4·(-3)·9
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{31}}{2*-3}=\frac{4-2\sqrt{31}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{31}}{2*-3}=\frac{4+2\sqrt{31}}{-6}
$