Solve 10x=4x^2-6

$$10 x = 4 x ^ { 2 } - 6$$

Solve for x

$$x=3$$

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

Steps Using Factoring By Grouping

Subtract 4x^{2} from both sides.

$$10x-4x^{2}=-6$$

Add 6 to both sides.

$$10x-4x^{2}+6=0$$

Divide both sides by 2.

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

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 4x^{2} from both sides.

$$10x-4x^{2}=-6$$

Add 6 to both sides.

$$10x-4x^{2}+6=0$$

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.

$$-4x^{2}+10x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 10 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-10±\sqrt{100+16\times 6}}{2\left(-4\right)}$$

Multiply 16 times 6.

$$x=\frac{-10±\sqrt{100+96}}{2\left(-4\right)}$$

Add 100 to 96.

$$x=\frac{-10±\sqrt{196}}{2\left(-4\right)}$$

Take the square root of 196.

$$x=\frac{-10±14}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-10±14}{-8}$$

Now solve the equation x=\frac{-10±14}{-8} when ± is plus. Add -10 to 14.

$$x=\frac{4}{-8}$$

Reduce the fraction \frac{4}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-10±14}{-8} when ± is minus. Subtract 14 from -10.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$10x-4x^{2}=-6$$

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.

$$-4x^{2}+10x=-6$$

Divide both sides by -4.

$$\frac{-4x^{2}+10x}{-4}=-\frac{6}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{10}{-4}x=-\frac{6}{-4}$$

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

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

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

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