Solve y^2-2y^2/10+y^2/4=1

$$\frac { y ^ { 2 } - 2 y ^ { 2 } } { 10 } + \frac { y ^ { 2 } } { 4 } = 1$$

Solve for y

$$y = \frac{2 \sqrt{15}}{3} \approx 2.581988897$$

$$y = -\frac{2 \sqrt{15}}{3} \approx -2.581988897$$

Steps by Finding Square Root

Multiply both sides of the equation by 20, the least common multiple of 10,4.

$$2\left(y^{2}-2y^{2}\right)+5y^{2}=20$$

Combine y^{2} and -2y^{2} to get -y^{2}.

$$2\left(-1\right)y^{2}+5y^{2}=20$$

Multiply 2 and -1 to get -2.

$$-2y^{2}+5y^{2}=20$$

Combine -2y^{2} and 5y^{2} to get 3y^{2}.

$$3y^{2}=20$$

Divide both sides by 3.

$$y^{2}=\frac{20}{3}$$

Take the square root of both sides of the equation.

$$y=\frac{2\sqrt{15}}{3} y=-\frac{2\sqrt{15}}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 20, the least common multiple of 10,4.

$$2\left(y^{2}-2y^{2}\right)+5y^{2}=20$$

Combine y^{2} and -2y^{2} to get -y^{2}.

$$2\left(-1\right)y^{2}+5y^{2}=20$$

Multiply 2 and -1 to get -2.

$$-2y^{2}+5y^{2}=20$$

Combine -2y^{2} and 5y^{2} to get 3y^{2}.

$$3y^{2}=20$$

Subtract 20 from both sides.

$$3y^{2}-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Square 0.

$$y=\frac{0±\sqrt{-4\times 3\left(-20\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{0±\sqrt{-12\left(-20\right)}}{2\times 3}$$

Multiply -12 times -20.

$$y=\frac{0±\sqrt{240}}{2\times 3}$$

Take the square root of 240.

$$y=\frac{0±4\sqrt{15}}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{0±4\sqrt{15}}{6}$$

Now solve the equation y=\frac{0±4\sqrt{15}}{6} when ± is plus.

$$y=\frac{2\sqrt{15}}{3}$$

Now solve the equation y=\frac{0±4\sqrt{15}}{6} when ± is minus.

$$y=-\frac{2\sqrt{15}}{3}$$

The equation is now solved.

$$y=\frac{2\sqrt{15}}{3} y=-\frac{2\sqrt{15}}{3}$$