Solve 3:x+4=x+6:x+14

$$3 : x + 4 = x + 6 : x + 14$$

Solve for x (complex solution)

$$x=\sqrt{22}-5\approx -0.30958424$$

$$x=-\left(\sqrt{22}+5\right)\approx -9.69041576$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3+x\times 4=xx+6+x\times 14$$

Multiply x and x to get x^{2}.

$$3+x\times 4=x^{2}+6+x\times 14$$

Subtract x^{2} from both sides.

$$3+x\times 4-x^{2}=6+x\times 14$$

Subtract 6 from both sides.

$$3+x\times 4-x^{2}-6=x\times 14$$

Subtract 6 from 3 to get -3.

$$-3+x\times 4-x^{2}=x\times 14$$

Subtract x\times 14 from both sides.

$$-3+x\times 4-x^{2}-x\times 14=0$$

Combine x\times 4 and -x\times 14 to get -10x.

$$-3-10x-x^{2}=0$$

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.

$$-x^{2}-10x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\left(-1\right)}$$

Add 100 to -12.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\left(-1\right)}$$

Take the square root of 88.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{22}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±2\sqrt{22}}{-2}$$

Now solve the equation x=\frac{10±2\sqrt{22}}{-2} when ± is plus. Add 10 to 2\sqrt{22}.

$$x=\frac{2\sqrt{22}+10}{-2}$$

Divide 10+2\sqrt{22} by -2.

$$x=-\left(\sqrt{22}+5\right)$$

Now solve the equation x=\frac{10±2\sqrt{22}}{-2} when ± is minus. Subtract 2\sqrt{22} from 10.

$$x=\frac{10-2\sqrt{22}}{-2}$$

Divide 10-2\sqrt{22} by -2.

$$x=\sqrt{22}-5$$

The equation is now solved.

$$x=-\left(\sqrt{22}+5\right) x=\sqrt{22}-5$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3+x\times 4=xx+6+x\times 14$$

Multiply x and x to get x^{2}.

$$3+x\times 4=x^{2}+6+x\times 14$$

Subtract x^{2} from both sides.

$$3+x\times 4-x^{2}=6+x\times 14$$

Subtract x\times 14 from both sides.

$$3+x\times 4-x^{2}-x\times 14=6$$

Combine x\times 4 and -x\times 14 to get -10x.

$$3-10x-x^{2}=6$$

Subtract 3 from both sides.

$$-10x-x^{2}=6-3$$

Subtract 3 from 6 to get 3.

$$-10x-x^{2}=3$$

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.

$$-x^{2}-10x=3$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{3}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=\frac{3}{-1}$$

Divide 3 by -1.

$$x^{2}+10x=-3$$

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.

$$x^{2}+10x+5^{2}=-3+5^{2}$$

Square 5.

$$x^{2}+10x+25=-3+25$$

Add -3 to 25.

$$x^{2}+10x+25=22$$

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}.

$$\left(x+5\right)^{2}=22$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{22}$$

Simplify.

$$x+5=\sqrt{22} x+5=-\sqrt{22}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{22}-5 x=-\sqrt{22}-5$$

Solve for x

$$x=\sqrt{22}-5\approx -0.30958424$$

$$x=-\sqrt{22}-5\approx -9.69041576$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3+x\times 4=xx+6+x\times 14$$

Multiply x and x to get x^{2}.

$$3+x\times 4=x^{2}+6+x\times 14$$

Subtract x^{2} from both sides.

$$3+x\times 4-x^{2}=6+x\times 14$$

Subtract 6 from both sides.

$$3+x\times 4-x^{2}-6=x\times 14$$

Subtract 6 from 3 to get -3.

$$-3+x\times 4-x^{2}=x\times 14$$

Subtract x\times 14 from both sides.

$$-3+x\times 4-x^{2}-x\times 14=0$$

Combine x\times 4 and -x\times 14 to get -10x.

$$-3-10x-x^{2}=0$$

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.

$$-x^{2}-10x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\left(-1\right)}$$

Add 100 to -12.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\left(-1\right)}$$

Take the square root of 88.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{22}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±2\sqrt{22}}{-2}$$

Now solve the equation x=\frac{10±2\sqrt{22}}{-2} when ± is plus. Add 10 to 2\sqrt{22}.

$$x=\frac{2\sqrt{22}+10}{-2}$$

Divide 10+2\sqrt{22} by -2.

$$x=-\left(\sqrt{22}+5\right)$$

Now solve the equation x=\frac{10±2\sqrt{22}}{-2} when ± is minus. Subtract 2\sqrt{22} from 10.

$$x=\frac{10-2\sqrt{22}}{-2}$$

Divide 10-2\sqrt{22} by -2.

$$x=\sqrt{22}-5$$

The equation is now solved.

$$x=-\left(\sqrt{22}+5\right) x=\sqrt{22}-5$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3+x\times 4=xx+6+x\times 14$$

Multiply x and x to get x^{2}.

$$3+x\times 4=x^{2}+6+x\times 14$$

Subtract x^{2} from both sides.

$$3+x\times 4-x^{2}=6+x\times 14$$

Subtract x\times 14 from both sides.

$$3+x\times 4-x^{2}-x\times 14=6$$

Combine x\times 4 and -x\times 14 to get -10x.

$$3-10x-x^{2}=6$$

Subtract 3 from both sides.

$$-10x-x^{2}=6-3$$

Subtract 3 from 6 to get 3.

$$-10x-x^{2}=3$$

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.

$$-x^{2}-10x=3$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{3}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=\frac{3}{-1}$$

Divide 3 by -1.

$$x^{2}+10x=-3$$

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.

$$x^{2}+10x+5^{2}=-3+5^{2}$$

Square 5.

$$x^{2}+10x+25=-3+25$$

Add -3 to 25.

$$x^{2}+10x+25=22$$

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}.

$$\left(x+5\right)^{2}=22$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{22}$$

Simplify.

$$x+5=\sqrt{22} x+5=-\sqrt{22}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{22}-5 x=-\sqrt{22}-5$$