Solve y^2=4y-3

$$y ^ { 2 } = 4 y - 3$$

Solve for y

$$y=1$$

$$y=3$$

Steps Using Factoring

Subtract 4y from both sides.

$$y^{2}-4y=-3$$

Add 3 to both sides.

$$y^{2}-4y+3=0$$

To solve the equation, factor y^{2}-4y+3 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-3\right)\left(y-1\right)$$

To find equation solutions, solve y-3=0 and y-1=0.

$$y=3 y=1$$

Steps Using Factoring By Grouping

Subtract 4y from both sides.

$$y^{2}-4y=-3$$

Add 3 to both sides.

$$y^{2}-4y+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite y^{2}-4y+3 as \left(y^{2}-3y\right)+\left(-y+3\right).

$$\left(y^{2}-3y\right)+\left(-y+3\right)$$

Factor out y in the first and -1 in the second group.

$$y\left(y-3\right)-\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(y-1\right)$$

To find equation solutions, solve y-3=0 and y-1=0.

$$y=3 y=1$$

Steps Using the Quadratic Formula

Subtract 4y from both sides.

$$y^{2}-4y=-3$$

Add 3 to both sides.

$$y^{2}-4y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2}$$

Multiply -4 times 3.

$$y=\frac{-\left(-4\right)±\sqrt{16-12}}{2}$$

Add 16 to -12.

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

Take the square root of 4.

$$y=\frac{-\left(-4\right)±2}{2}$$

The opposite of -4 is 4.

$$y=\frac{4±2}{2}$$

Now solve the equation y=\frac{4±2}{2} when ± is plus. Add 4 to 2.

$$y=\frac{6}{2}$$

Divide 6 by 2.

$$y=3$$

Now solve the equation y=\frac{4±2}{2} when ± is minus. Subtract 2 from 4.

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

Divide 2 by 2.

$$y=1$$

The equation is now solved.

$$y=3 y=1$$

Steps for Completing the Square

Subtract 4y from both sides.

$$y^{2}-4y=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-4y+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$y^{2}-4y+4=-3+4$$

Add -3 to 4.

$$y^{2}-4y+4=1$$

Factor y^{2}-4y+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$y-2=1 y-2=-1$$

Add 2 to both sides of the equation.

$$y=3 y=1$$