Solve 10x^2+26x-3=x^2

$$10 x ^ { 2 } + 26 x - 3 = x ^ { 2 }$$

Solve for x

$$x=-3$$

$$x=\frac{1}{9}\approx 0.111111111$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$10x^{2}+26x-3-x^{2}=0$$

Combine 10x^{2} and -x^{2} to get 9x^{2}.

$$9x^{2}+26x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=26 ab=9\left(-3\right)=-27$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -27.

$$-1,27 -3,9$$

Calculate the sum for each pair.

$$-1+27=26 -3+9=6$$

The solution is the pair that gives sum 26.

$$a=-1 b=27$$

Rewrite 9x^{2}+26x-3 as \left(9x^{2}-x\right)+\left(27x-3\right).

$$\left(9x^{2}-x\right)+\left(27x-3\right)$$

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

$$x\left(9x-1\right)+3\left(9x-1\right)$$

Factor out common term 9x-1 by using distributive property.

$$\left(9x-1\right)\left(x+3\right)$$

To find equation solutions, solve 9x-1=0 and x+3=0.

$$x=\frac{1}{9} x=-3$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$10x^{2}+26x-3-x^{2}=0$$

Combine 10x^{2} and -x^{2} to get 9x^{2}.

$$9x^{2}+26x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 26 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\times 9\left(-3\right)}}{2\times 9}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 9\left(-3\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-26±\sqrt{676-36\left(-3\right)}}{2\times 9}$$

Multiply -36 times -3.

$$x=\frac{-26±\sqrt{676+108}}{2\times 9}$$

Add 676 to 108.

$$x=\frac{-26±\sqrt{784}}{2\times 9}$$

Take the square root of 784.

$$x=\frac{-26±28}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-26±28}{18}$$

Now solve the equation x=\frac{-26±28}{18} when ± is plus. Add -26 to 28.

$$x=\frac{2}{18}$$

Reduce the fraction \frac{2}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{9}$$

Now solve the equation x=\frac{-26±28}{18} when ± is minus. Subtract 28 from -26.

$$x=-\frac{54}{18}$$

Divide -54 by 18.

$$x=-3$$

The equation is now solved.

$$x=\frac{1}{9} x=-3$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$10x^{2}+26x-3-x^{2}=0$$

Combine 10x^{2} and -x^{2} to get 9x^{2}.

$$9x^{2}+26x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$9x^{2}+26x=3$$

Divide both sides by 9.

$$\frac{9x^{2}+26x}{9}=\frac{3}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{26}{9}x=\frac{3}{9}$$

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

$$x^{2}+\frac{26}{9}x=\frac{1}{3}$$

Divide \frac{26}{9}, the coefficient of the x term, by 2 to get \frac{13}{9}. Then add the square of \frac{13}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{26}{9}x+\left(\frac{13}{9}\right)^{2}=\frac{1}{3}+\left(\frac{13}{9}\right)^{2}$$

Square \frac{13}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{26}{9}x+\frac{169}{81}=\frac{1}{3}+\frac{169}{81}$$

Add \frac{1}{3} to \frac{169}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{26}{9}x+\frac{169}{81}=\frac{196}{81}$$

Factor x^{2}+\frac{26}{9}x+\frac{169}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{9}\right)^{2}=\frac{196}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{9}\right)^{2}}=\sqrt{\frac{196}{81}}$$

Simplify.

$$x+\frac{13}{9}=\frac{14}{9} x+\frac{13}{9}=-\frac{14}{9}$$

Subtract \frac{13}{9} from both sides of the equation.

$$x=\frac{1}{9} x=-3$$