Solve 4x^2=-5x+6

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

Solve for x

$$x=-2$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Add 5x to both sides.

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

Subtract 6 from both sides.

$$4x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=4\left(-6\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite 4x^{2}+5x-6 as \left(4x^{2}-3x\right)+\left(8x-6\right).

$$\left(4x^{2}-3x\right)+\left(8x-6\right)$$

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

$$x\left(4x-3\right)+2\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(x+2\right)$$

To find equation solutions, solve 4x-3=0 and x+2=0.

$$x=\frac{3}{4} x=-2$$

Steps Using the Quadratic Formula

Add 5x to both sides.

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

Subtract 6 from both sides.

$$4x^{2}+5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-6\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-6\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-6\right)}}{2\times 4}$$

Multiply -16 times -6.

$$x=\frac{-5±\sqrt{25+96}}{2\times 4}$$

Add 25 to 96.

$$x=\frac{-5±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$x=\frac{-5±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±11}{8}$$

Now solve the equation x=\frac{-5±11}{8} when ± is plus. Add -5 to 11.

$$x=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-5±11}{8} when ± is minus. Subtract 11 from -5.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=\frac{3}{4} x=-2$$

Steps for Completing the Square

Add 5x to both sides.

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

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=\frac{6}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{4}x=\frac{3}{2}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{3}{2}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{3}{2}+\frac{25}{64}$$

Add \frac{3}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{121}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{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}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{11}{8} x+\frac{5}{8}=-\frac{11}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{3}{4} x=-2$$