What is the answer to this quadratic: 0=−2a2±2b−12 −2a2+2b−12=0

Step 1
${\displaystyle 2a^2+2b-12=0}$ is not a solvable quadratic. I will assume you meant
${\displaystyle 2x^2+2x-12}$
The quadratic formula is:
${\displaystyle \frac{-b\pm \sqrt{b^2-\left(4ac\right)}}{2a}}$
${\displaystyle \frac{-2\pm \sqrt{2^2-(4\times 2-12)}}{2\times 2}}$
${\displaystyle \frac{-2\pm \sqrt{4-(-96)}}{4}}$
${\displaystyle \frac{-2\pm \sqrt{100}}{4}}$
${\displaystyle \frac{-2\pm 10}{4}}$
${\displaystyle \frac{12}{4}}$ or ${\displaystyle \frac{8}{4}}$
The answer is ${\displaystyle x=3}$ or ${\displaystyle x=2}$

Step 1
Given equation: ${\displaystyle 2a^2+2b-12=0}$
Quadratic equations like this one, with an ${\displaystyle x^2}$ term but no x term, can still be solved using the quadratic formula, ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$, once they are put in standard form: ${\displaystyle a{x}^2+bx+c=0}$
This equation is in standard form: ${\displaystyle a{x}^2+bx+c=0}$
Substitute 2 for a, 0 fro b, and ${\displaystyle -12+2b}$ for c in the quadratic formula, ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
${\displaystyle a=\frac{0\pm \sqrt{0^2-4\times 2(2b-12)}}{2\times 2}}$
Square 0
${\displaystyle a=\frac{0\pm \sqrt{-4\times 2(2b-12)}}{2\times 2}}$
Multiply -4 times 2
${\displaystyle a=\frac{0\pm \sqrt{-8(2b-12)}}{2\times 2}}$
Multiply -8 times ${\displaystyle -12+2b}$
${\displaystyle a=\frac{0\pm \sqrt{96-16b}}{2\times 2}}$
Take the square root of ${\displaystyle 96-16b}$
${\displaystyle a=\frac{0\pm 4\sqrt{6-b}}{2\times 2}}$
Multiply 2 times 2
${\displaystyle a=\frac{0\pm 4\sqrt{6-b}}{4}}$
Now solve the equation ${\displaystyle a=\frac{0\pm 4\sqrt{6-b}}{4}}$ when ${\displaystyle \pm }$ is plus and minus
${\displaystyle a=\sqrt{6-b}}$
${\displaystyle a=-\sqrt{6-b}}$

Step 1
$2a^2+2b-12=0$
Subtract $2a^2$ from both sides. Anything subtracted from zero gives its negation.
$2b-12=-2a^2$
Add 12 to both sides
$2b=-2a^2+12$
The equation is in standard form.
$2b=12-2a^2$
Divide both sides by 2
$\frac{2b}{2}=\frac{12-2a^2}{2}$
Dividing by 2 undoes the multiplication by 2.
$b=\frac{12-2a^2}{2}$
Divide $-2a^2+12$ by 2
$b=6-a^2$