Solve - 5y ^ 2 = 4y ^ 2 - 25 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$-5y^{2}=4y^{2}-25$$

Solve for y

$y = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$
$y = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Subtract $4y^{2}$ from both sides.

$$-5y^{2}-4y^{2}=-25$$

Combine $-5y^{2}$ and $-4y^{2}$ to get $-9y^{2}$.

$$-9y^{2}=-25$$

Divide both sides by $-9$.

$$y^{2}=\frac{-25}{-9}$$

Fraction $\frac{-25}{-9}$ can be simplified to $\frac{25}{9}$ by removing the negative sign from both the numerator and the denominator.

$$y^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$y=\frac{5}{3}$$ $$y=-\frac{5}{3}$$

Steps Using the Quadratic Formula

Subtract $4y^{2}$ from both sides.

$$-5y^{2}-4y^{2}=-25$$

Combine $-5y^{2}$ and $-4y^{2}$ to get $-9y^{2}$.

$$-9y^{2}=-25$$

Add $25$ to both sides.

$$-9y^{2}+25=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-9$ for $a$, $0$ for $b$, and $25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{0±\sqrt{0^{2}-4\left(-9\right)\times 25}}{2\left(-9\right)}$$

Square $0$.

$$y=\frac{0±\sqrt{-4\left(-9\right)\times 25}}{2\left(-9\right)}$$

Multiply $-4$ times $-9$.

$$y=\frac{0±\sqrt{36\times 25}}{2\left(-9\right)}$$

Multiply $36$ times $25$.

$$y=\frac{0±\sqrt{900}}{2\left(-9\right)}$$

Take the square root of $900$.

$$y=\frac{0±30}{2\left(-9\right)}$$

Multiply $2$ times $-9$.

$$y=\frac{0±30}{-18}$$

Now solve the equation $y=\frac{0±30}{-18}$ when $±$ is plus. Reduce the fraction $\frac{30}{-18}$ to lowest terms by extracting and canceling out $6$.

$$y=-\frac{5}{3}$$

Now solve the equation $y=\frac{0±30}{-18}$ when $±$ is minus. Reduce the fraction $\frac{-30}{-18}$ to lowest terms by extracting and canceling out $6$.

$$y=\frac{5}{3}$$

The equation is now solved.

$$y=-\frac{5}{3}$$ $$y=\frac{5}{3}$$