Solve 7y-2=5y^2

$$7 y - 2 = 5 y ^ { 2 }$$

Solve for y

$$y=1$$

$$y=\frac{2}{5}=0.4$$

Steps Using Factoring By Grouping

Subtract 5y^{2} from both sides.

$$7y-2-5y^{2}=0$$

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

$$-5y^{2}+7y-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5y^{2}+ay+by-2. To find a and b, set up a system to be solved.

$$a+b=7 ab=-5\left(-2\right)=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=5 b=2$$

Rewrite -5y^{2}+7y-2 as \left(-5y^{2}+5y\right)+\left(2y-2\right).

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

Factor out 5y in the first and -2 in the second group.

$$5y\left(-y+1\right)-2\left(-y+1\right)$$

Factor out common term -y+1 by using distributive property.

$$\left(-y+1\right)\left(5y-2\right)$$

To find equation solutions, solve -y+1=0 and 5y-2=0.

$$y=1 y=\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 5y^{2} from both sides.

$$7y-2-5y^{2}=0$$

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.

$$-5y^{2}+7y-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 7 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-7±\sqrt{7^{2}-4\left(-5\right)\left(-2\right)}}{2\left(-5\right)}$$

Square 7.

$$y=\frac{-7±\sqrt{49-4\left(-5\right)\left(-2\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-7±\sqrt{49+20\left(-2\right)}}{2\left(-5\right)}$$

Multiply 20 times -2.

$$y=\frac{-7±\sqrt{49-40}}{2\left(-5\right)}$$

Add 49 to -40.

$$y=\frac{-7±\sqrt{9}}{2\left(-5\right)}$$

Take the square root of 9.

$$y=\frac{-7±3}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{-7±3}{-10}$$

Now solve the equation y=\frac{-7±3}{-10} when ± is plus. Add -7 to 3.

$$y=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-7±3}{-10} when ± is minus. Subtract 3 from -7.

$$y=-\frac{10}{-10}$$

Divide -10 by -10.

$$y=1$$

The equation is now solved.

$$y=\frac{2}{5} y=1$$

Steps for Completing the Square

Subtract 5y^{2} from both sides.

$$7y-2-5y^{2}=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$7y-5y^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5y^{2}+7y=2$$

Divide both sides by -5.

$$\frac{-5y^{2}+7y}{-5}=\frac{2}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$y^{2}+\frac{7}{-5}y=\frac{2}{-5}$$

Divide 7 by -5.

$$y^{2}-\frac{7}{5}y=\frac{2}{-5}$$

Divide 2 by -5.

$$y^{2}-\frac{7}{5}y=-\frac{2}{5}$$

Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{7}{5}y+\left(-\frac{7}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{7}{10}\right)^{2}$$

Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{7}{5}y+\frac{49}{100}=-\frac{2}{5}+\frac{49}{100}$$

Add -\frac{2}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{7}{5}y+\frac{49}{100}=\frac{9}{100}$$

Factor y^{2}-\frac{7}{5}y+\frac{49}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{7}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

$$y-\frac{7}{10}=\frac{3}{10} y-\frac{7}{10}=-\frac{3}{10}$$

Add \frac{7}{10} to both sides of the equation.

$$y=1 y=\frac{2}{5}$$