Solve 10x^2-5x-5+1=3x

$$10 x ^ { 2 } - 5 x - 5 + 1 = 3 x$$

Solve for x

$$x = \frac{\sqrt{14} + 2}{5} \approx 1.148331477$$

$$x=\frac{2-\sqrt{14}}{5}\approx -0.348331477$$

Steps Using the Quadratic Formula

Add -5 and 1 to get -4.

$$10x^{2}-5x-4=3x$$

Subtract 3x from both sides.

$$10x^{2}-5x-4-3x=0$$

Combine -5x and -3x to get -8x.

$$10x^{2}-8x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 10\left(-4\right)}}{2\times 10}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 10\left(-4\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-8\right)±\sqrt{64-40\left(-4\right)}}{2\times 10}$$

Multiply -40 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+160}}{2\times 10}$$

Add 64 to 160.

$$x=\frac{-\left(-8\right)±\sqrt{224}}{2\times 10}$$

Take the square root of 224.

$$x=\frac{-\left(-8\right)±4\sqrt{14}}{2\times 10}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{14}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{8±4\sqrt{14}}{20}$$

Now solve the equation x=\frac{8±4\sqrt{14}}{20} when ± is plus. Add 8 to 4\sqrt{14}.

$$x=\frac{4\sqrt{14}+8}{20}$$

Divide 8+4\sqrt{14} by 20.

$$x=\frac{\sqrt{14}+2}{5}$$

Now solve the equation x=\frac{8±4\sqrt{14}}{20} when ± is minus. Subtract 4\sqrt{14} from 8.

$$x=\frac{8-4\sqrt{14}}{20}$$

Divide 8-4\sqrt{14} by 20.

$$x=\frac{2-\sqrt{14}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{14}+2}{5} x=\frac{2-\sqrt{14}}{5}$$

Steps for Completing the Square

Add -5 and 1 to get -4.

$$10x^{2}-5x-4=3x$$

Subtract 3x from both sides.

$$10x^{2}-5x-4-3x=0$$

Combine -5x and -3x to get -8x.

$$10x^{2}-8x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$10x^{2}-8x=4$$

Divide both sides by 10.

$$\frac{10x^{2}-8x}{10}=\frac{4}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{8}{10}\right)x=\frac{4}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{5}x=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{5}x=\frac{2}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{2}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{2}{5}+\frac{4}{25}$$

Add \frac{2}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{14}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{14}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{\sqrt{14}}{5} x-\frac{2}{5}=-\frac{\sqrt{14}}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{\sqrt{14}+2}{5} x=\frac{2-\sqrt{14}}{5}$$