Solve 2x^2-8x*10=14

$$2 x ^ { 2 } - 8 x \cdot 10 = 14$$

Solve for x

$$x=\sqrt{407}+20\approx 40.174241002$$

$$x=20-\sqrt{407}\approx -0.174241002$$

Steps Using the Quadratic Formula

Multiply 8 and 10 to get 80.

$$2x^{2}-80x=14$$

Subtract 14 from both sides.

$$2x^{2}-80x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -80 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 2\left(-14\right)}}{2\times 2}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 2\left(-14\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-80\right)±\sqrt{6400-8\left(-14\right)}}{2\times 2}$$

Multiply -8 times -14.

$$x=\frac{-\left(-80\right)±\sqrt{6400+112}}{2\times 2}$$

Add 6400 to 112.

$$x=\frac{-\left(-80\right)±\sqrt{6512}}{2\times 2}$$

Take the square root of 6512.

$$x=\frac{-\left(-80\right)±4\sqrt{407}}{2\times 2}$$

The opposite of -80 is 80.

$$x=\frac{80±4\sqrt{407}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{80±4\sqrt{407}}{4}$$

Now solve the equation x=\frac{80±4\sqrt{407}}{4} when ± is plus. Add 80 to 4\sqrt{407}.

$$x=\frac{4\sqrt{407}+80}{4}$$

Divide 80+4\sqrt{407} by 4.

$$x=\sqrt{407}+20$$

Now solve the equation x=\frac{80±4\sqrt{407}}{4} when ± is minus. Subtract 4\sqrt{407} from 80.

$$x=\frac{80-4\sqrt{407}}{4}$$

Divide 80-4\sqrt{407} by 4.

$$x=20-\sqrt{407}$$

The equation is now solved.

$$x=\sqrt{407}+20 x=20-\sqrt{407}$$

Steps for Completing the Square

Multiply 8 and 10 to get 80.

$$2x^{2}-80x=14$$

Divide both sides by 2.

$$\frac{2x^{2}-80x}{2}=\frac{14}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{80}{2}\right)x=\frac{14}{2}$$

Divide -80 by 2.

$$x^{2}-40x=\frac{14}{2}$$

Divide 14 by 2.

$$x^{2}-40x=7$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=7+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=7+400$$

Add 7 to 400.

$$x^{2}-40x+400=407$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=407$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{407}$$

Simplify.

$$x-20=\sqrt{407} x-20=-\sqrt{407}$$

Add 20 to both sides of the equation.

$$x=\sqrt{407}+20 x=20-\sqrt{407}$$