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

$$( - x + 1 - 2 \times 4 x ) = ( 3 x ( - 5 ) - 4 x ( 3 x )$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3}i}{12}-\frac{1}{4}\approx -0.25+0.144337567i$$

$$x=-\frac{\sqrt{3}i}{12}-\frac{1}{4}\approx -0.25-0.144337567i$$

Steps Using the Quadratic Formula

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

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

Multiply 3 and -5 to get -15.

$$-x+1-2\times 4x=-15x-4x^{2}\times 3$$

Multiply 4 and 3 to get 12.

$$-x+1-2\times 4x=-15x-12x^{2}$$

Add 15x to both sides.

$$-x+1-2\times 4x+15x=-12x^{2}$$

Add 12x^{2} to both sides.

$$-x+1-2\times 4x+15x+12x^{2}=0$$

Combine -x and 15x to get 14x.

$$14x+1-2\times 4x+12x^{2}=0$$

Combine all terms containing x.

$$\left(14-2\times 4\right)x+1+12x^{2}=0$$

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.

$$12x^{2}+6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 12}}{2\times 12}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 12}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-6±\sqrt{36-48}}{2\times 12}$$

Add 36 to -48.

$$x=\frac{-6±\sqrt{-12}}{2\times 12}$$

Take the square root of -12.

$$x=\frac{-6±2\sqrt{3}i}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-6±2\sqrt{3}i}{24}$$

Now solve the equation x=\frac{-6±2\sqrt{3}i}{24} when ± is plus. Add -6 to 2i\sqrt{3}.

$$x=\frac{-6+2\sqrt{3}i}{24}$$

Divide -6+2i\sqrt{3} by 24.

$$x=\frac{\sqrt{3}i}{12}-\frac{1}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{3}i}{24} when ± is minus. Subtract 2i\sqrt{3} from -6.

$$x=\frac{-2\sqrt{3}i-6}{24}$$

Divide -6-2i\sqrt{3} by 24.

$$x=-\frac{\sqrt{3}i}{12}-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{3}i}{12}-\frac{1}{4} x=-\frac{\sqrt{3}i}{12}-\frac{1}{4}$$

Steps for Completing the Square

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

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

Multiply 3 and -5 to get -15.

$$-x+1-2\times 4x=-15x-4x^{2}\times 3$$

Multiply 4 and 3 to get 12.

$$-x+1-2\times 4x=-15x-12x^{2}$$

Add 15x to both sides.

$$-x+1-2\times 4x+15x=-12x^{2}$$

Add 12x^{2} to both sides.

$$-x+1-2\times 4x+15x+12x^{2}=0$$

Combine -x and 15x to get 14x.

$$14x+1-2\times 4x+12x^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$14x-2\times 4x+12x^{2}=-1$$

Combine all terms containing x.

$$\left(14-2\times 4\right)x+12x^{2}=-1$$

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.

$$12x^{2}+6x=-1$$

Divide both sides by 12.

$$\frac{12x^{2}+6x}{12}=-\frac{1}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{6}{12}x=-\frac{1}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{1}{2}x=-\frac{1}{12}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{1}{12}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{1}{12}+\frac{1}{16}$$

Add -\frac{1}{12} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{1}{48}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{1}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{1}{48}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{3}i}{12} x+\frac{1}{4}=-\frac{\sqrt{3}i}{12}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{3}i}{12}-\frac{1}{4} x=-\frac{\sqrt{3}i}{12}-\frac{1}{4}$$