Solve 5x+3=2x^2

$$5 x + 3 = 2 x ^ { 2 }$$

Solve for x

$$x=3$$

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

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$5x+3-2x^{2}=0$$

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

$$-2x^{2}+5x+3=0$$

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.

$$a+b=5 ab=-2\times 3=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

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

The solution is the pair that gives sum 5.

$$a=6 b=-1$$

Rewrite -2x^{2}+5x+3 as \left(-2x^{2}+6x\right)+\left(-x+3\right).

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

Factor out 2x in -2x^{2}+6x.

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

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.

$$5x+3-2x^{2}=0$$

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.

$$-2x^{2}+5x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-5±\sqrt{25+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-5±\sqrt{25+24}}{2\left(-2\right)}$$

Add 25 to 24.

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

Take the square root of 49.

$$x=\frac{-5±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-5±7}{-4}$$

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

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

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{-5±7}{-4} when ± is minus. Subtract 7 from -5.

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

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.

$$5x+3-2x^{2}=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$5x-2x^{2}=-3$$

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.

$$-2x^{2}+5x=-3$$

Divide both sides by -2.

$$\frac{-2x^{2}+5x}{-2}=-\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{5}{-2}x=-\frac{3}{-2}$$

Divide 5 by -2.

$$x^{2}-\frac{5}{2}x=-\frac{3}{-2}$$

Divide -3 by -2.

$$x^{2}-\frac{5}{2}x=\frac{3}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

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

Add \frac{5}{4} to both sides of the equation.

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