Solve for x

x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667

$x=35 =132 ≈1.666666667$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Direct Factoring Method

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Graph in 2D

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

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

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

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=-226 -3-75=-78 -5-45=-50 -9-25=-34 -15-15=-30

Calculate the sum for each pair.

a=-15 b=-15

The solution is the pair that gives sum -30.

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

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

3x\left(3x-5\right)-5\left(3x-5\right)

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

\left(3x-5\right)\left(3x-5\right)

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

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

Rewrite as a binomial square.

x=\frac{5}{3}

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

9x^{2}-30x+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.

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

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}.

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

Square -30.

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

Multiply -4 times 9.

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

Multiply -36 times 25.

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

Add 900 to -900.

x=-\frac{-30}{2\times 9}

Take the square root of 0.

x=\frac{30}{2\times 9}

The opposite of -30 is 30.

x=\frac{30}{18}

Multiply 2 times 9.

x=\frac{5}{3}

Reduce the fraction \frac{30}{18}\approx 1.666666667 to lowest terms by extracting and canceling out 6.

9x^{2}-30x+25=0

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.

9x^{2}-30x+25-25=-25

Subtract 25 from both sides of the equation.

9x^{2}-30x=-25

Subtracting 25 from itself leaves 0.

\frac{9x^{2}-30x}{9}=\frac{-25}{9}

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

Reduce the fraction \frac{-30}{9}\approx -3.333333333 to lowest terms by extracting and canceling out 3.

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

Divide -25 by 9.

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

Divide -\frac{10}{3}\approx -3.333333333, the coefficient of the x term, by 2 to get -\frac{5}{3}\approx -1.666666667. Then add the square of -\frac{5}{3}\approx -1.666666667 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{3}\approx -1.666666667 by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{25}{9}\approx -2.777777778 to \frac{25}{9}\approx 2.777777778 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Take the square root of both sides of the equation.

x-\frac{5}{3}=0 x-\frac{5}{3}=0

Simplify.

x=\frac{5}{3} x=\frac{5}{3}

Add \frac{5}{3}\approx 1.666666667 to both sides of the equation.

x=\frac{5}{3}

The equation is now solved. Solutions are the same.

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

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 9

r + s = \frac{10}{3} rs = \frac{25}{9}

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = \frac{5}{3} - u s = \frac{5}{3} + u

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(\frac{5}{3} - u) (\frac{5}{3} + u) = \frac{25}{9}

To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{9}

\frac{25}{9} - u^2 = \frac{25}{9}

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = \frac{25}{9}-\frac{25}{9} = 0

Simplify the expression by subtracting \frac{25}{9} on both sides

u^2 = 0 u = 0

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r = s = \frac{5}{3} = 1.667

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

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