Solve -4x+6=x^2+6x+10

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

Solve for x (complex solution)

$$x=\sqrt{21}-5\approx -0.417424305$$

$$x=-\left(\sqrt{21}+5\right)\approx -9.582575695$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine -4x and -6x to get -10x.

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

Subtract 10 from both sides.

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

Subtract 10 from 6 to get -4.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16}}{2\left(-1\right)}$$

Add 100 to -16.

$$x=\frac{-\left(-10\right)±\sqrt{84}}{2\left(-1\right)}$$

Take the square root of 84.

$$x=\frac{-\left(-10\right)±2\sqrt{21}}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±2\sqrt{21}}{-2}$$

Now solve the equation x=\frac{10±2\sqrt{21}}{-2} when ± is plus. Add 10 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}+10}{-2}$$

Divide 10+2\sqrt{21} by -2.

$$x=-\left(\sqrt{21}+5\right)$$

Now solve the equation x=\frac{10±2\sqrt{21}}{-2} when ± is minus. Subtract 2\sqrt{21} from 10.

$$x=\frac{10-2\sqrt{21}}{-2}$$

Divide 10-2\sqrt{21} by -2.

$$x=\sqrt{21}-5$$

The equation is now solved.

$$x=-\left(\sqrt{21}+5\right) x=\sqrt{21}-5$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine -4x and -6x to get -10x.

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

Subtract 6 from both sides.

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

Subtract 6 from 10 to get 4.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{4}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=\frac{4}{-1}$$

Divide 4 by -1.

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

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-4+5^{2}$$

Square 5.

$$x^{2}+10x+25=-4+25$$

Add -4 to 25.

$$x^{2}+10x+25=21$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{21}$$

Simplify.

$$x+5=\sqrt{21} x+5=-\sqrt{21}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$

Solve for x

$$x=\sqrt{21}-5\approx -0.417424305$$

$$x=-\sqrt{21}-5\approx -9.582575695$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine -4x and -6x to get -10x.

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

Subtract 10 from both sides.

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

Subtract 10 from 6 to get -4.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16}}{2\left(-1\right)}$$

Add 100 to -16.

$$x=\frac{-\left(-10\right)±\sqrt{84}}{2\left(-1\right)}$$

Take the square root of 84.

$$x=\frac{-\left(-10\right)±2\sqrt{21}}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±2\sqrt{21}}{-2}$$

Now solve the equation x=\frac{10±2\sqrt{21}}{-2} when ± is plus. Add 10 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}+10}{-2}$$

Divide 10+2\sqrt{21} by -2.

$$x=-\left(\sqrt{21}+5\right)$$

Now solve the equation x=\frac{10±2\sqrt{21}}{-2} when ± is minus. Subtract 2\sqrt{21} from 10.

$$x=\frac{10-2\sqrt{21}}{-2}$$

Divide 10-2\sqrt{21} by -2.

$$x=\sqrt{21}-5$$

The equation is now solved.

$$x=-\left(\sqrt{21}+5\right) x=\sqrt{21}-5$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine -4x and -6x to get -10x.

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

Subtract 6 from both sides.

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

Subtract 6 from 10 to get 4.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{4}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=\frac{4}{-1}$$

Divide 4 by -1.

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

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-4+5^{2}$$

Square 5.

$$x^{2}+10x+25=-4+25$$

Add -4 to 25.

$$x^{2}+10x+25=21$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{21}$$

Simplify.

$$x+5=\sqrt{21} x+5=-\sqrt{21}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$