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Start by isolating $3x^2$ alone on one side by subtracting $9$ from both sides. We can then isolate the $x^2$-term by dividing both sides by $3.$ As the last step, we will square root both sides in order to solve for $x.$

$3x^2+9=9$

SubEqn$\text{LHS}-9=\text{RHS}-9$

$3x^2=0$

DivEqn$\left.\text{LHS}\middle/3\right.=\left.\text{RHS}\middle/3\right.$

$x^2=0$

SqrtEqn$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$x=\pm \sqrt{0}$

SimpRootSimplify root

$x=0$

$x=0$ solves the equation. Note that $x=\text{-}0$ and $x=+0$ are equal, and so there is only one solution: $x=0.$

b

On the left side, we have two identical terms, i.e. $x^2.$ This means the sum can be written as the product $2x^2.$ By dividing both sides by $2,$ you get $x^2$ alone on one side, and can then square root both sides.

$x^2+x^2=200$

SumToProdTwoTerms$a+a=2a$

$2x^2=200$

DivEqn$\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.$

$x^2=100$

SqrtEqn$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$x=\pm\sqrt{100}$

SimpRootSimplify root

$x=\pm 10$

Both $x=\text{-}10$ and $x=10$ are solutions to the equation.

c

Start by isolating $x^2$ on one side. Then take the square root of both sides.

$2x^2-2=126$

AddEqn$\text{LHS}+2=\text{RHS}+2$

$2x^2=128$

DivEqn$\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.$

$x^2=64$

SqrtEqn$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$x=\pm \sqrt{64}$

SimpRootSimplify root

$x=\pm 8$

Both $x=\text{-}8$ and $x=8$ are solutions to the equation.