x^2+2x+5=4x+14

Solution for x^2+2x+5=4x+14 equation:

x^2+2x+5=4x+14

We move all terms to the left:

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

We get rid of parentheses

x^2+2x-4x-14+5=0

We add all the numbers together, and all the variables

x^2-2x-9=0

a = 1; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·1·(-9)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{10}}{2*1}=\frac{2-2\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*1}=\frac{2+2\sqrt{10}}{2}
$