X^2+19+10=-12x+28

Solution for X^2+19+10=-12x+28 equation:

X^2+19+10=-12X+28

We move all terms to the left:

X^2+19+10-(-12X+28)=0

We add all the numbers together, and all the variables

X^2-(-12X+28)+29=0

We get rid of parentheses

X^2+12X-28+29=0

We add all the numbers together, and all the variables

X^2+12X+1=0

a = 1; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·1·1
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{35}}{2*1}=\frac{-12-2\sqrt{35}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{35}}{2*1}=\frac{-12+2\sqrt{35}}{2}
$