14x+17=2(x+3)(x+3)-5

Solution for 14x+17=2(x+3)(x+3)-5 equation:

14x+17=2(x+3)(x+3)-5

We move all terms to the left:

14x+17-(2(x+3)(x+3)-5)=0

We multiply parentheses ..

-(2(+x^2+3x+3x+9)-5)+14x+17=0


We calculate terms in parentheses: -(2(+x^2+3x+3x+9)-5), so:

2(+x^2+3x+3x+9)-5

We multiply parentheses

2x^2+6x+6x+18-5

We add all the numbers together, and all the variables

2x^2+12x+13

Back to the equation:

-(2x^2+12x+13)


We add all the numbers together, and all the variables

14x-(2x^2+12x+13)+17=0

We get rid of parentheses

-2x^2+14x-12x-13+17=0

We add all the numbers together, and all the variables

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

a = -2; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-2}=\frac{4}{-4}
=-1
$