Solve 8+9y^2=17

$$8 + 9 y ^ { 2 } = 17$$

Solve for y

$$y=1$$

$$y=-1$$

Steps Using Difference of Squares

Subtract 17 from both sides.

$$8+9y^{2}-17=0$$

Subtract 17 from 8 to get -9.

$$-9+9y^{2}=0$$

Divide both sides by 9.

$$-1+y^{2}=0$$

Consider -1+y^{2}. Rewrite -1+y^{2} as y^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(y-1\right)\left(y+1\right)=0$$

To find equation solutions, solve y-1=0 and y+1=0.

$$y=1 y=-1$$

Steps by Finding Square Root

Subtract 8 from both sides.

$$9y^{2}=17-8$$

Subtract 8 from 17 to get 9.

$$9y^{2}=9$$

Divide both sides by 9.

$$y^{2}=\frac{9}{9}$$

Divide 9 by 9 to get 1.

$$y^{2}=1$$

Take the square root of both sides of the equation.

$$y=1 y=-1$$

Steps Using the Quadratic Formula

Subtract 17 from both sides.

$$8+9y^{2}-17=0$$

Subtract 17 from 8 to get -9.

$$-9+9y^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$9y^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 9\left(-9\right)}}{2\times 9}$$

Square 0.

$$y=\frac{0±\sqrt{-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{0±\sqrt{-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$y=\frac{0±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$y=\frac{0±18}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{0±18}{18}$$

Now solve the equation y=\frac{0±18}{18} when ± is plus. Divide 18 by 18.

$$y=1$$

Now solve the equation y=\frac{0±18}{18} when ± is minus. Divide -18 by 18.

$$y=-1$$

The equation is now solved.

$$y=1 y=-1$$