Solve -(3x-4)4(x-5)=2(7-4x)

$$-(3x-4)4(x-5)=2(7-4x)$$

Solve for x

$$x=\frac{\sqrt{159}}{6}+\frac{7}{2}\approx 5.601586702$$

$$x=-\frac{\sqrt{159}}{6}+\frac{7}{2}\approx 1.398413298$$

Steps Using the Quadratic Formula

To find the opposite of 3x-4, find the opposite of each term.

$$\left(-3x-\left(-4\right)\right)\times 4\left(x-5\right)=2\left(7-4x\right)$$

The opposite of -4 is 4.

$$\left(-3x+4\right)\times 4\left(x-5\right)=2\left(7-4x\right)$$

Use the distributive property to multiply -3x+4 by 4.

$$\left(-12x+16\right)\left(x-5\right)=2\left(7-4x\right)$$

Apply the distributive property by multiplying each term of -12x+16 by each term of x-5.

$$-12x^{2}+60x+16x-80=2\left(7-4x\right)$$

Combine 60x and 16x to get 76x.

$$-12x^{2}+76x-80=2\left(7-4x\right)$$

Use the distributive property to multiply 2 by 7-4x.

$$-12x^{2}+76x-80=14-8x$$

Subtract 14 from both sides.

$$-12x^{2}+76x-80-14=-8x$$

Subtract 14 from -80 to get -94.

$$-12x^{2}+76x-94=-8x$$

Add 8x to both sides.

$$-12x^{2}+76x-94+8x=0$$

Combine 76x and 8x to get 84x.

$$-12x^{2}+84x-94=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 84 for b, and -94 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-84±\sqrt{84^{2}-4\left(-12\right)\left(-94\right)}}{2\left(-12\right)}$$

Square 84.

$$x=\frac{-84±\sqrt{7056-4\left(-12\right)\left(-94\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-84±\sqrt{7056+48\left(-94\right)}}{2\left(-12\right)}$$

Multiply 48 times -94.

$$x=\frac{-84±\sqrt{7056-4512}}{2\left(-12\right)}$$

Add 7056 to -4512.

$$x=\frac{-84±\sqrt{2544}}{2\left(-12\right)}$$

Take the square root of 2544.

$$x=\frac{-84±4\sqrt{159}}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-84±4\sqrt{159}}{-24}$$

Now solve the equation x=\frac{-84±4\sqrt{159}}{-24} when ± is plus. Add -84 to 4\sqrt{159}.

$$x=\frac{4\sqrt{159}-84}{-24}$$

Divide -84+4\sqrt{159} by -24.

$$x=-\frac{\sqrt{159}}{6}+\frac{7}{2}$$

Now solve the equation x=\frac{-84±4\sqrt{159}}{-24} when ± is minus. Subtract 4\sqrt{159} from -84.

$$x=\frac{-4\sqrt{159}-84}{-24}$$

Divide -84-4\sqrt{159} by -24.

$$x=\frac{\sqrt{159}}{6}+\frac{7}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{159}}{6}+\frac{7}{2} x=\frac{\sqrt{159}}{6}+\frac{7}{2}$$

Steps for Completing the Square

To find the opposite of 3x-4, find the opposite of each term.

$$\left(-3x-\left(-4\right)\right)\times 4\left(x-5\right)=2\left(7-4x\right)$$

The opposite of -4 is 4.

$$\left(-3x+4\right)\times 4\left(x-5\right)=2\left(7-4x\right)$$

Use the distributive property to multiply -3x+4 by 4.

$$\left(-12x+16\right)\left(x-5\right)=2\left(7-4x\right)$$

Apply the distributive property by multiplying each term of -12x+16 by each term of x-5.

$$-12x^{2}+60x+16x-80=2\left(7-4x\right)$$

Combine 60x and 16x to get 76x.

$$-12x^{2}+76x-80=2\left(7-4x\right)$$

Use the distributive property to multiply 2 by 7-4x.

$$-12x^{2}+76x-80=14-8x$$

Add 8x to both sides.

$$-12x^{2}+76x-80+8x=14$$

Combine 76x and 8x to get 84x.

$$-12x^{2}+84x-80=14$$

Add 80 to both sides.

$$-12x^{2}+84x=14+80$$

Add 14 and 80 to get 94.

$$-12x^{2}+84x=94$$

Divide both sides by -12.

$$\frac{-12x^{2}+84x}{-12}=\frac{94}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{84}{-12}x=\frac{94}{-12}$$

Divide 84 by -12.

$$x^{2}-7x=\frac{94}{-12}$$

Reduce the fraction \frac{94}{-12} to lowest terms by extracting and canceling out 2.

$$x^{2}-7x=-\frac{47}{6}$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-\frac{47}{6}+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-\frac{47}{6}+\frac{49}{4}$$

Add -\frac{47}{6} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-7x+\frac{49}{4}=\frac{53}{12}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{53}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{53}{12}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{159}}{6} x-\frac{7}{2}=-\frac{\sqrt{159}}{6}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{159}}{6}+\frac{7}{2} x=-\frac{\sqrt{159}}{6}+\frac{7}{2}$$