Solve y^2+12=7y

$$y ^ { 2 } + 12 = 7 y$$

Solve for y

$$y=3$$

$$y=4$$

Steps Using Factoring

Subtract 7y from both sides.

$$y^{2}+12-7y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-7y+12=0$$

To solve the equation, factor y^{2}-7y+12 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.

$$a+b=-7 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

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

$$\left(y-4\right)\left(y-3\right)$$

To find equation solutions, solve y-4=0 and y-3=0.

$$y=4 y=3$$

Steps Using Factoring By Grouping

Subtract 7y from both sides.

$$y^{2}+12-7y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-7y+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite y^{2}-7y+12 as \left(y^{2}-4y\right)+\left(-3y+12\right).

$$\left(y^{2}-4y\right)+\left(-3y+12\right)$$

Factor out y in the first and -3 in the second group.

$$y\left(y-4\right)-3\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y-3\right)$$

To find equation solutions, solve y-4=0 and y-3=0.

$$y=4 y=3$$

Steps Using the Quadratic Formula

Subtract 7y from both sides.

$$y^{2}+12-7y=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y^{2}-7y+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Square -7.

$$y=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$y=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$y=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-\left(-7\right)±1}{2}$$

The opposite of -7 is 7.

$$y=\frac{7±1}{2}$$

Now solve the equation y=\frac{7±1}{2} when ± is plus. Add 7 to 1.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

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

Divide 6 by 2.

$$y=3$$

The equation is now solved.

$$y=4 y=3$$

Steps for Completing the Square

Subtract 7y from both sides.

$$y^{2}+12-7y=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-7y=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-7y+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-7y+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$y^{2}-7y+\frac{49}{4}=\frac{1}{4}$$

Factor y^{2}-7y+\frac{49}{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}.

$$\left(y-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y-\frac{7}{2}=\frac{1}{2} y-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$y=4 y=3$$