(x+3)(2x-5)=14x-8

Solution for (x+3)(2x-5)=14x-8 equation:

(x+3)(2x-5)=14x-8

We move all terms to the left:

(x+3)(2x-5)-(14x-8)=0

We get rid of parentheses

(x+3)(2x-5)-14x+8=0

We multiply parentheses ..

(+2x^2-5x+6x-15)-14x+8=0

We get rid of parentheses

2x^2-5x+6x-14x-15+8=0

We add all the numbers together, and all the variables

2x^2-13x-7=0

a = 2; b = -13; c = -7;
Δ = b2-4ac
Δ = -132-4·2·(-7)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-15}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+15}{2*2}=\frac{28}{4}
=7
$