Solve 7x-3=2x^2

$$7 x - 3 = 2 x ^ { 2 }$$

Solve for x

$$x=3$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$7x-3-2x^{2}=0$$

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

$$-2x^{2}+7x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\left(-3\right)=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=6 b=1$$

Rewrite -2x^{2}+7x-3 as \left(-2x^{2}+6x\right)+\left(x-3\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(-x+3\right)-\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x-1\right)$$

To find equation solutions, solve -x+3=0 and 2x-1=0.

$$x=3 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$7x-3-2x^{2}=0$$

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.

$$-2x^{2}+7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-7±\sqrt{49+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-7±\sqrt{49-24}}{2\left(-2\right)}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-7±5}{-4}$$

Now solve the equation x=\frac{-7±5}{-4} when ± is plus. Add -7 to 5.

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

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

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

Now solve the equation x=\frac{-7±5}{-4} when ± is minus. Subtract 5 from -7.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

The equation is now solved.

$$x=\frac{1}{2} x=3$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$7x-3-2x^{2}=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$7x-2x^{2}=3$$

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.

$$-2x^{2}+7x=3$$

Divide both sides by -2.

$$\frac{-2x^{2}+7x}{-2}=\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{7}{-2}x=\frac{3}{-2}$$

Divide 7 by -2.

$$x^{2}-\frac{7}{2}x=\frac{3}{-2}$$

Divide 3 by -2.

$$x^{2}-\frac{7}{2}x=-\frac{3}{2}$$

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.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{7}{4}\right)^{2}$$

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

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{5}{4} x-\frac{7}{4}=-\frac{5}{4}$$

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

$$x=3 x=\frac{1}{2}$$