Solve 2(1-(2x-4))-(4x^2+3x)=5-4*(x+2)-6=

$$2(1-(2x-4))-(4 { x }^{ 2 } +3x)=5-4 \times (x+2)-6=$$

Solve for x

$$x = \frac{\sqrt{313} - 3}{8} \approx 1.836475752$$

$$x=\frac{-\sqrt{313}-3}{8}\approx -2.586475752$$

Steps Using the Quadratic Formula

To find the opposite of 2x-4, find the opposite of each term.

$$2\left(1-2x+4\right)-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

Add 1 and 4 to get 5.

$$2\left(5-2x\right)-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

Use the distributive property to multiply 2 by 5-2x.

$$10-4x-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

To find the opposite of 4x^{2}+3x, find the opposite of each term.

$$10-4x-4x^{2}-3x=5-4\left(x+2\right)-6$$

Combine -4x and -3x to get -7x.

$$10-7x-4x^{2}=5-4\left(x+2\right)-6$$

Use the distributive property to multiply -4 by x+2.

$$10-7x-4x^{2}=5-4x-8-6$$

Subtract 8 from 5 to get -3.

$$10-7x-4x^{2}=-3-4x-6$$

Subtract 6 from -3 to get -9.

$$10-7x-4x^{2}=-9-4x$$

Subtract -9 from both sides.

$$10-7x-4x^{2}-\left(-9\right)=-4x$$

The opposite of -9 is 9.

$$10-7x-4x^{2}+9=-4x$$

Add 4x to both sides.

$$10-7x-4x^{2}+9+4x=0$$

Add 10 and 9 to get 19.

$$19-7x-4x^{2}+4x=0$$

Combine -7x and 4x to get -3x.

$$19-3x-4x^{2}=0$$

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.

$$-4x^{2}-3x+19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -3 for b, and 19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)\times 19}}{2\left(-4\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)\times 19}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16\times 19}}{2\left(-4\right)}$$

Multiply 16 times 19.

$$x=\frac{-\left(-3\right)±\sqrt{9+304}}{2\left(-4\right)}$$

Add 9 to 304.

$$x=\frac{-\left(-3\right)±\sqrt{313}}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{313}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{3±\sqrt{313}}{-8}$$

Now solve the equation x=\frac{3±\sqrt{313}}{-8} when ± is plus. Add 3 to \sqrt{313}.

$$x=\frac{\sqrt{313}+3}{-8}$$

Divide 3+\sqrt{313} by -8.

$$x=\frac{-\sqrt{313}-3}{8}$$

Now solve the equation x=\frac{3±\sqrt{313}}{-8} when ± is minus. Subtract \sqrt{313} from 3.

$$x=\frac{3-\sqrt{313}}{-8}$$

Divide 3-\sqrt{313} by -8.

$$x=\frac{\sqrt{313}-3}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{313}-3}{8} x=\frac{\sqrt{313}-3}{8}$$

Steps for Completing the Square

To find the opposite of 2x-4, find the opposite of each term.

$$2\left(1-2x+4\right)-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

Add 1 and 4 to get 5.

$$2\left(5-2x\right)-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

Use the distributive property to multiply 2 by 5-2x.

$$10-4x-\left(4x^{2}+3x\right)=5-4\left(x+2\right)-6$$

To find the opposite of 4x^{2}+3x, find the opposite of each term.

$$10-4x-4x^{2}-3x=5-4\left(x+2\right)-6$$

Combine -4x and -3x to get -7x.

$$10-7x-4x^{2}=5-4\left(x+2\right)-6$$

Use the distributive property to multiply -4 by x+2.

$$10-7x-4x^{2}=5-4x-8-6$$

Subtract 8 from 5 to get -3.

$$10-7x-4x^{2}=-3-4x-6$$

Subtract 6 from -3 to get -9.

$$10-7x-4x^{2}=-9-4x$$

Add 4x to both sides.

$$10-7x-4x^{2}+4x=-9$$

Combine -7x and 4x to get -3x.

$$10-3x-4x^{2}=-9$$

Subtract 10 from both sides.

$$-3x-4x^{2}=-9-10$$

Subtract 10 from -9 to get -19.

$$-3x-4x^{2}=-19$$

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.

$$-4x^{2}-3x=-19$$

Divide both sides by -4.

$$\frac{-4x^{2}-3x}{-4}=-\frac{19}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{3}{-4}\right)x=-\frac{19}{-4}$$

Divide -3 by -4.

$$x^{2}+\frac{3}{4}x=-\frac{19}{-4}$$

Divide -19 by -4.

$$x^{2}+\frac{3}{4}x=\frac{19}{4}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{19}{4}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{19}{4}+\frac{9}{64}$$

Add \frac{19}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{313}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{313}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{313}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{\sqrt{313}}{8} x+\frac{3}{8}=-\frac{\sqrt{313}}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{\sqrt{313}-3}{8} x=\frac{-\sqrt{313}-3}{8}$$