3x^2+43=30x-5

Solution for 3x^2+43=30x-5 equation:

3x^2+43=30x-5

We move all terms to the left:

3x^2+43-(30x-5)=0

We get rid of parentheses

3x^2-30x+5+43=0

We add all the numbers together, and all the variables

3x^2-30x+48=0

a = 3; b = -30; c = +48;
Δ = b2-4ac
Δ = -302-4·3·48
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-18}{2*3}=\frac{12}{6}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+18}{2*3}=\frac{48}{6}
=8
$