Solve 7x-1-5-2x=4x-3-1+4x^2

$$7 x - 1 - 5 - 2 x = 4 x - 3 - 1 + 4 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{31}i+1}{8}\approx 0.125-0.695970545i$$

$$x=\frac{1+\sqrt{31}i}{8}\approx 0.125+0.695970545i$$

Steps Using the Quadratic Formula

Subtract 5 from -1 to get -6.

$$7x-6-2x=4x-3-1+4x^{2}$$

Combine 7x and -2x to get 5x.

$$5x-6=4x-3-1+4x^{2}$$

Subtract 1 from -3 to get -4.

$$5x-6=4x-4+4x^{2}$$

Subtract 4x from both sides.

$$5x-6-4x=-4+4x^{2}$$

Combine 5x and -4x to get x.

$$x-6=-4+4x^{2}$$

Subtract -4 from both sides.

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

The opposite of -4 is 4.

$$x-6+4=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x-6+4-4x^{2}=0$$

Add -6 and 4 to get -2.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -4.

$$x=\frac{-1±\sqrt{1+16\left(-2\right)}}{2\left(-4\right)}$$

Multiply 16 times -2.

$$x=\frac{-1±\sqrt{1-32}}{2\left(-4\right)}$$

Add 1 to -32.

$$x=\frac{-1±\sqrt{-31}}{2\left(-4\right)}$$

Take the square root of -31.

$$x=\frac{-1±\sqrt{31}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-1±\sqrt{31}i}{-8}$$

Now solve the equation x=\frac{-1±\sqrt{31}i}{-8} when ± is plus. Add -1 to i\sqrt{31}.

$$x=\frac{-1+\sqrt{31}i}{-8}$$

Divide -1+i\sqrt{31} by -8.

$$x=\frac{-\sqrt{31}i+1}{8}$$

Now solve the equation x=\frac{-1±\sqrt{31}i}{-8} when ± is minus. Subtract i\sqrt{31} from -1.

$$x=\frac{-\sqrt{31}i-1}{-8}$$

Divide -1-i\sqrt{31} by -8.

$$x=\frac{1+\sqrt{31}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{31}i+1}{8} x=\frac{1+\sqrt{31}i}{8}$$

Steps for Completing the Square

Subtract 5 from -1 to get -6.

$$7x-6-2x=4x-3-1+4x^{2}$$

Combine 7x and -2x to get 5x.

$$5x-6=4x-3-1+4x^{2}$$

Subtract 1 from -3 to get -4.

$$5x-6=4x-4+4x^{2}$$

Subtract 4x from both sides.

$$5x-6-4x=-4+4x^{2}$$

Combine 5x and -4x to get x.

$$x-6=-4+4x^{2}$$

Subtract 4x^{2} from both sides.

$$x-6-4x^{2}=-4$$

Add 6 to both sides.

$$x-4x^{2}=-4+6$$

Add -4 and 6 to get 2.

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

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.

$$-4x^{2}+x=2$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{1}{-4}x=\frac{2}{-4}$$

Divide 1 by -4.

$$x^{2}-\frac{1}{4}x=\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{4}x=-\frac{1}{2}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=-\frac{1}{2}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{1}{2}+\frac{1}{64}$$

Add -\frac{1}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{31}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{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}.

$$\left(x-\frac{1}{8}\right)^{2}=-\frac{31}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{-\frac{31}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{31}i}{8} x-\frac{1}{8}=-\frac{\sqrt{31}i}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i+1}{8}$$