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

Solution for (2x^2+3x+5)+(x-7)(4x+3)=0 equation:

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

We get rid of parentheses

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

We multiply parentheses ..

2x^2+(+4x^2+3x-28x-21)+3x+5=0

We get rid of parentheses

2x^2+4x^2+3x-28x+3x-21+5=0

We add all the numbers together, and all the variables

6x^2-22x-16=0

a = 6; b = -22; c = -16;
Δ = b2-4ac
Δ = -222-4·6·(-16)
Δ = 868
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{217}}{2*6}=\frac{22-2\sqrt{217}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{217}}{2*6}=\frac{22+2\sqrt{217}}{12}
$