Solve 3x^2-35=45-2x^2

$$3 x ^ { 2 } - 35 = 45 - 2 x ^ { 2 }$$

Solve for x

$$x=4$$

$$x=-4$$

Steps Using Difference of Squares

Subtract 45 from both sides.

$$3x^{2}-35-45=-2x^{2}$$

Subtract 45 from -35 to get -80.

$$3x^{2}-80=-2x^{2}$$

Add 2x^{2} to both sides.

$$3x^{2}-80+2x^{2}=0$$

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$5x^{2}-80=0$$

Divide both sides by 5.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Steps by Finding Square Root

Add 2x^{2} to both sides.

$$3x^{2}-35+2x^{2}=45$$

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$5x^{2}-35=45$$

Add 35 to both sides.

$$5x^{2}=45+35$$

Add 45 and 35 to get 80.

$$5x^{2}=80$$

Divide both sides by 5.

$$x^{2}=\frac{80}{5}$$

Divide 80 by 5 to get 16.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Subtract 45 from both sides.

$$3x^{2}-35-45=-2x^{2}$$

Subtract 45 from -35 to get -80.

$$3x^{2}-80=-2x^{2}$$

Add 2x^{2} to both sides.

$$3x^{2}-80+2x^{2}=0$$

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$5x^{2}-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-80\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-80\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-80\right)}}{2\times 5}$$

Multiply -20 times -80.

$$x=\frac{0±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$x=\frac{0±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±40}{10}$$

Now solve the equation x=\frac{0±40}{10} when ± is plus. Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{0±40}{10} when ± is minus. Divide -40 by 10.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$