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

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

Solve for x

$$x=1$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$7x-1-2x^{2}=4$$

Subtract 4 from both sides.

$$7x-1-2x^{2}-4=0$$

Subtract 4 from -1 to get -5.

$$7x-5-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+7x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\left(-5\right)=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=5 b=2$$

Rewrite -2x^{2}+7x-5 as \left(-2x^{2}+5x\right)+\left(2x-5\right).

$$\left(-2x^{2}+5x\right)+\left(2x-5\right)$$

Factor out -x in -2x^{2}+5x.

$$-x\left(2x-5\right)+2x-5$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(-x+1\right)$$

To find equation solutions, solve 2x-5=0 and -x+1=0.

$$x=\frac{5}{2} x=1$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$7x-1-2x^{2}=4$$

Subtract 4 from both sides.

$$7x-1-2x^{2}-4=0$$

Subtract 4 from -1 to get -5.

$$7x-5-2x^{2}=0$$

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.

$$-2x^{2}+7x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 7 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-7±\sqrt{49+8\left(-5\right)}}{2\left(-2\right)}$$

Multiply 8 times -5.

$$x=\frac{-7±\sqrt{49-40}}{2\left(-2\right)}$$

Add 49 to -40.

$$x=\frac{-7±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of 9.

$$x=\frac{-7±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-7±3}{-4}$$

Now solve the equation x=\frac{-7±3}{-4} when ± is plus. Add -7 to 3.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-7±3}{-4} when ± is minus. Subtract 3 from -7.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=1 x=\frac{5}{2}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$7x-1-2x^{2}=4$$

Add 1 to both sides.

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

Add 4 and 1 to get 5.

$$7x-2x^{2}=5$$

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.

$$-2x^{2}+7x=5$$

Divide both sides by -2.

$$\frac{-2x^{2}+7x}{-2}=\frac{5}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{7}{-2}x=\frac{5}{-2}$$

Divide 7 by -2.

$$x^{2}-\frac{7}{2}x=\frac{5}{-2}$$

Divide 5 by -2.

$$x^{2}-\frac{7}{2}x=-\frac{5}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{5}{2}+\frac{49}{16}$$

Add -\frac{5}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{3}{4} x-\frac{7}{4}=-\frac{3}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{5}{2} x=1$$