(7x^2)+5x+3=(4x^2)+14x+33

Solution for (7x^2)+5x+3=(4x^2)+14x+33 equation:

(7x^2)+5x+3=(4x^2)+14x+33

We move all terms to the left:

(7x^2)+5x+3-((4x^2)+14x+33)=0

We get rid of parentheses

7x^2-4x^2+5x-14x-33+3=0

We add all the numbers together, and all the variables

3x^2-9x-30=0

a = 3; b = -9; c = -30;
Δ = b2-4ac
Δ = -92-4·3·(-30)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-21}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+21}{2*3}=\frac{30}{6}
=5
$