Solve -40y+28y^2

$$- 40 y + 28 y ^ { 2 }$$

Factor

$$4y\left(7y-10\right)$$

Solution Steps

Factor out 4.

$$4\left(-10y+7y^{2}\right)$$

Consider -10y+7y^{2}. Factor out y.

$$y\left(-10+7y\right)$$

Rewrite the complete factored expression.

$$4y\left(7y-10\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$28y^{2}-40y=0$$

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.

$$y=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}}}{2\times 28}$$

Take the square root of \left(-40\right)^{2}.

$$y=\frac{-\left(-40\right)±40}{2\times 28}$$

The opposite of -40 is 40.

$$y=\frac{40±40}{2\times 28}$$

Multiply 2 times 28.

$$y=\frac{40±40}{56}$$

Now solve the equation y=\frac{40±40}{56} when ± is plus. Add 40 to 40.

$$y=\frac{80}{56}$$

Reduce the fraction \frac{80}{56} to lowest terms by extracting and canceling out 8.

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

Now solve the equation y=\frac{40±40}{56} when ± is minus. Subtract 40 from 40.

$$y=\frac{0}{56}$$

Divide 0 by 56.

$$y=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{10}{7} for x_{1} and 0 for x_{2}.

$$28y^{2}-40y=28\left(y-\frac{10}{7}\right)y$$

Subtract \frac{10}{7} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$28y^{2}-40y=28\times \frac{7y-10}{7}y$$

Cancel out 7, the greatest common factor in 28 and 7.

$$28y^{2}-40y=4\left(7y-10\right)y$$

Evaluate

$$4y\left(7y-10\right)$$