Solve 30v=9v^2+25

$$30 v = 9 v ^ { 2 } + 25$$

Solve for v

$$v = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Subtract 9v^{2} from both sides.

$$30v-9v^{2}=25$$

Subtract 25 from both sides.

$$30v-9v^{2}-25=0$$

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

$$-9v^{2}+30v-25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9v^{2}+av+bv-25. To find a and b, set up a system to be solved.

$$a+b=30 ab=-9\left(-25\right)=225$$

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 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 30.

$$a=15 b=15$$

Rewrite -9v^{2}+30v-25 as \left(-9v^{2}+15v\right)+\left(15v-25\right).

$$\left(-9v^{2}+15v\right)+\left(15v-25\right)$$

Factor out -3v in the first and 5 in the second group.

$$-3v\left(3v-5\right)+5\left(3v-5\right)$$

Factor out common term 3v-5 by using distributive property.

$$\left(3v-5\right)\left(-3v+5\right)$$

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

$$v=\frac{5}{3} v=\frac{5}{3}$$

Steps Using the Quadratic Formula

Subtract 9v^{2} from both sides.

$$30v-9v^{2}=25$$

Subtract 25 from both sides.

$$30v-9v^{2}-25=0$$

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.

$$-9v^{2}+30v-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 30 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-30±\sqrt{30^{2}-4\left(-9\right)\left(-25\right)}}{2\left(-9\right)}$$

Square 30.

$$v=\frac{-30±\sqrt{900-4\left(-9\right)\left(-25\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$v=\frac{-30±\sqrt{900+36\left(-25\right)}}{2\left(-9\right)}$$

Multiply 36 times -25.

$$v=\frac{-30±\sqrt{900-900}}{2\left(-9\right)}$$

Add 900 to -900.

$$v=\frac{-30±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$v=-\frac{30}{2\left(-9\right)}$$

Multiply 2 times -9.

$$v=-\frac{30}{-18}$$

Reduce the fraction \frac{-30}{-18} to lowest terms by extracting and canceling out 6.

$$v=\frac{5}{3}$$

Steps for Completing the Square

Subtract 9v^{2} from both sides.

$$30v-9v^{2}=25$$

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.

$$-9v^{2}+30v=25$$

Divide both sides by -9.

$$\frac{-9v^{2}+30v}{-9}=\frac{25}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$v^{2}+\frac{30}{-9}v=\frac{25}{-9}$$

Reduce the fraction \frac{30}{-9} to lowest terms by extracting and canceling out 3.

$$v^{2}-\frac{10}{3}v=\frac{25}{-9}$$

Divide 25 by -9.

$$v^{2}-\frac{10}{3}v=-\frac{25}{9}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-\frac{10}{3}v+\left(-\frac{5}{3}\right)^{2}=-\frac{25}{9}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-\frac{10}{3}v+\frac{25}{9}=\frac{-25+25}{9}$$

Add -\frac{25}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v^{2}-\frac{10}{3}v+\frac{25}{9}=0$$

Factor v^{2}-\frac{10}{3}v+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-\frac{5}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{5}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$v-\frac{5}{3}=0 v-\frac{5}{3}=0$$

Add \frac{5}{3} to both sides of the equation.

$$v=\frac{5}{3} v=\frac{5}{3}$$

The equation is now solved. Solutions are the same.

$$v=\frac{5}{3}$$