Solve 4x*2x+3x=72

$$4x \times 2x+3x=72$$

Solve for x

$$x = \frac{3 \sqrt{257} - 3}{16} \approx 2.818353664$$

$$x=\frac{-3\sqrt{257}-3}{16}\approx -3.193353664$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$4x^{2}\times 2+3x=72$$

Multiply 4 and 2 to get 8.

$$8x^{2}+3x=72$$

Subtract 72 from both sides.

$$8x^{2}+3x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 3 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 8\left(-72\right)}}{2\times 8}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 8\left(-72\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-3±\sqrt{9-32\left(-72\right)}}{2\times 8}$$

Multiply -32 times -72.

$$x=\frac{-3±\sqrt{9+2304}}{2\times 8}$$

Add 9 to 2304.

$$x=\frac{-3±\sqrt{2313}}{2\times 8}$$

Take the square root of 2313.

$$x=\frac{-3±3\sqrt{257}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-3±3\sqrt{257}}{16}$$

Now solve the equation x=\frac{-3±3\sqrt{257}}{16} when ± is plus. Add -3 to 3\sqrt{257}.

$$x=\frac{3\sqrt{257}-3}{16}$$

Now solve the equation x=\frac{-3±3\sqrt{257}}{16} when ± is minus. Subtract 3\sqrt{257} from -3.

$$x=\frac{-3\sqrt{257}-3}{16}$$

The equation is now solved.

$$x=\frac{3\sqrt{257}-3}{16} x=\frac{-3\sqrt{257}-3}{16}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$4x^{2}\times 2+3x=72$$

Multiply 4 and 2 to get 8.

$$8x^{2}+3x=72$$

Divide both sides by 8.

$$\frac{8x^{2}+3x}{8}=\frac{72}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{3}{8}x=\frac{72}{8}$$

Divide 72 by 8.

$$x^{2}+\frac{3}{8}x=9$$

Divide \frac{3}{8}, the coefficient of the x term, by 2 to get \frac{3}{16}. Then add the square of \frac{3}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{8}x+\left(\frac{3}{16}\right)^{2}=9+\left(\frac{3}{16}\right)^{2}$$

Square \frac{3}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=9+\frac{9}{256}$$

Add 9 to \frac{9}{256}.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{2313}{256}$$

Factor x^{2}+\frac{3}{8}x+\frac{9}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{16}\right)^{2}=\frac{2313}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{16}\right)^{2}}=\sqrt{\frac{2313}{256}}$$

Simplify.

$$x+\frac{3}{16}=\frac{3\sqrt{257}}{16} x+\frac{3}{16}=-\frac{3\sqrt{257}}{16}$$

Subtract \frac{3}{16} from both sides of the equation.

$$x=\frac{3\sqrt{257}-3}{16} x=\frac{-3\sqrt{257}-3}{16}$$