Solve 9w^2+25=30w

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

Solve for w

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

Steps Using Factoring By Grouping

Subtract 30w from both sides.

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

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

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

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

$$a+b=-30 ab=9\times 25=225$$

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

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

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

$$3w\left(3w-5\right)-5\left(3w-5\right)$$

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

$$\left(3w-5\right)\left(3w-5\right)$$

Rewrite as a binomial square.

$$\left(3w-5\right)^{2}$$

To find equation solution, solve 3w-5=0.

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

Steps Using the Quadratic Formula

Subtract 30w from both sides.

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

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.

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

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}.

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

Square -30.

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

Multiply -4 times 9.

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

Multiply -36 times 25.

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

Add 900 to -900.

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

Take the square root of 0.

$$w=-\frac{-30}{2\times 9}$$

The opposite of -30 is 30.

$$w=\frac{30}{2\times 9}$$

Multiply 2 times 9.

$$w=\frac{30}{18}$$

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

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

Steps for Completing the Square

Subtract 30w from both sides.

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

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

$$w^{2}+\left(-\frac{30}{9}\right)w=-\frac{25}{9}$$

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

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

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.

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

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

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

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.

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

Factor w^{2}-\frac{10}{3}w+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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

The equation is now solved. Solutions are the same.

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