(2x2+-5x+-3)+(x+-3)=

Solution for (2x2+-5x+-3)+(x+-3)= equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-4x-6=0

a = 2; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*2}=\frac{-4}{4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*2}=\frac{12}{4}
=3
$