Solve 2y^2+6/7+y=3

$$\frac { 2 y ^ { 2 } + 6 } { 7 } + y = 3$$

Solve for y

$$y=-5$$

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 7.

$$2y^{2}+6+7y=21$$

Subtract 21 from both sides.

$$2y^{2}+6+7y-21=0$$

Subtract 21 from 6 to get -15.

$$2y^{2}-15+7y=0$$

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

$$2y^{2}+7y-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2y^{2}+ay+by-15. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\left(-15\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 2y^{2}+7y-15 as \left(2y^{2}-3y\right)+\left(10y-15\right).

$$\left(2y^{2}-3y\right)+\left(10y-15\right)$$

Factor out y in the first and 5 in the second group.

$$y\left(2y-3\right)+5\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(y+5\right)$$

To find equation solutions, solve 2y-3=0 and y+5=0.

$$y=\frac{3}{2} y=-5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 7.

$$2y^{2}+6+7y=21$$

Subtract 21 from both sides.

$$2y^{2}+6+7y-21=0$$

Subtract 21 from 6 to get -15.

$$2y^{2}-15+7y=0$$

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.

$$2y^{2}+7y-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-7±\sqrt{7^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square 7.

$$y=\frac{-7±\sqrt{49-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-7±\sqrt{49-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$y=\frac{-7±\sqrt{49+120}}{2\times 2}$$

Add 49 to 120.

$$y=\frac{-7±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$y=\frac{-7±13}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-7±13}{4}$$

Now solve the equation y=\frac{-7±13}{4} when ± is plus. Add -7 to 13.

$$y=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-7±13}{4} when ± is minus. Subtract 13 from -7.

$$y=-\frac{20}{4}$$

Divide -20 by 4.

$$y=-5$$

The equation is now solved.

$$y=\frac{3}{2} y=-5$$

Steps for Completing the Square

Multiply both sides of the equation by 7.

$$2y^{2}+6+7y=21$$

Subtract 6 from both sides.

$$2y^{2}+7y=21-6$$

Subtract 6 from 21 to get 15.

$$2y^{2}+7y=15$$

Divide both sides by 2.

$$\frac{2y^{2}+7y}{2}=\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\frac{7}{2}y=\frac{15}{2}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{7}{2}y+\left(\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{7}{2}y+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{7}{2}y+\frac{49}{16}=\frac{169}{16}$$

Factor y^{2}+\frac{7}{2}y+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$y+\frac{7}{4}=\frac{13}{4} y+\frac{7}{4}=-\frac{13}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$y=\frac{3}{2} y=-5$$