Solve 13x+(1+2x)(1-2x)=61x+50(x+3)

$$13 x + ( 1 + 2 x ) ( 1 - 2 x ) = 61 x + 50 ( x + 3 )$$

Solve for x

$$x=\frac{19\sqrt{5}-49}{4}\approx -1.628677107$$

$$x=\frac{-19\sqrt{5}-49}{4}\approx -22.871322893$$

Steps Using the Quadratic Formula

Consider \left(1+2x\right)\left(1-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$13x+1-\left(2x\right)^{2}=61x+50\left(x+3\right)$$

Expand \left(2x\right)^{2}.

$$13x+1-2^{2}x^{2}=61x+50\left(x+3\right)$$

Calculate 2 to the power of 2 and get 4.

$$13x+1-4x^{2}=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$13x+1-4x^{2}=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$13x+1-4x^{2}=111x+150$$

Subtract 111x from both sides.

$$13x+1-4x^{2}-111x=150$$

Combine 13x and -111x to get -98x.

$$-98x+1-4x^{2}=150$$

Subtract 150 from both sides.

$$-98x+1-4x^{2}-150=0$$

Subtract 150 from 1 to get -149.

$$-98x-149-4x^{2}=0$$

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.

$$-4x^{2}-98x-149=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -98 for b, and -149 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-98\right)±\sqrt{\left(-98\right)^{2}-4\left(-4\right)\left(-149\right)}}{2\left(-4\right)}$$

Square -98.

$$x=\frac{-\left(-98\right)±\sqrt{9604-4\left(-4\right)\left(-149\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-98\right)±\sqrt{9604+16\left(-149\right)}}{2\left(-4\right)}$$

Multiply 16 times -149.

$$x=\frac{-\left(-98\right)±\sqrt{9604-2384}}{2\left(-4\right)}$$

Add 9604 to -2384.

$$x=\frac{-\left(-98\right)±\sqrt{7220}}{2\left(-4\right)}$$

Take the square root of 7220.

$$x=\frac{-\left(-98\right)±38\sqrt{5}}{2\left(-4\right)}$$

The opposite of -98 is 98.

$$x=\frac{98±38\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{98±38\sqrt{5}}{-8}$$

Now solve the equation x=\frac{98±38\sqrt{5}}{-8} when ± is plus. Add 98 to 38\sqrt{5}.

$$x=\frac{38\sqrt{5}+98}{-8}$$

Divide 98+38\sqrt{5} by -8.

$$x=\frac{-19\sqrt{5}-49}{4}$$

Now solve the equation x=\frac{98±38\sqrt{5}}{-8} when ± is minus. Subtract 38\sqrt{5} from 98.

$$x=\frac{98-38\sqrt{5}}{-8}$$

Divide 98-38\sqrt{5} by -8.

$$x=\frac{19\sqrt{5}-49}{4}$$

The equation is now solved.

$$x=\frac{-19\sqrt{5}-49}{4} x=\frac{19\sqrt{5}-49}{4}$$

Steps for Completing the Square

Consider \left(1+2x\right)\left(1-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$13x+1-\left(2x\right)^{2}=61x+50\left(x+3\right)$$

Expand \left(2x\right)^{2}.

$$13x+1-2^{2}x^{2}=61x+50\left(x+3\right)$$

Calculate 2 to the power of 2 and get 4.

$$13x+1-4x^{2}=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$13x+1-4x^{2}=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$13x+1-4x^{2}=111x+150$$

Subtract 111x from both sides.

$$13x+1-4x^{2}-111x=150$$

Combine 13x and -111x to get -98x.

$$-98x+1-4x^{2}=150$$

Subtract 1 from both sides.

$$-98x-4x^{2}=150-1$$

Subtract 1 from 150 to get 149.

$$-98x-4x^{2}=149$$

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.

$$-4x^{2}-98x=149$$

Divide both sides by -4.

$$\frac{-4x^{2}-98x}{-4}=\frac{149}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{98}{-4}\right)x=\frac{149}{-4}$$

Reduce the fraction \frac{-98}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{49}{2}x=\frac{149}{-4}$$

Divide 149 by -4.

$$x^{2}+\frac{49}{2}x=-\frac{149}{4}$$

Divide \frac{49}{2}, the coefficient of the x term, by 2 to get \frac{49}{4}. Then add the square of \frac{49}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{49}{2}x+\left(\frac{49}{4}\right)^{2}=-\frac{149}{4}+\left(\frac{49}{4}\right)^{2}$$

Square \frac{49}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{49}{2}x+\frac{2401}{16}=-\frac{149}{4}+\frac{2401}{16}$$

Add -\frac{149}{4} to \frac{2401}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{49}{2}x+\frac{2401}{16}=\frac{1805}{16}$$

Factor x^{2}+\frac{49}{2}x+\frac{2401}{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}.

$$\left(x+\frac{49}{4}\right)^{2}=\frac{1805}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{49}{4}\right)^{2}}=\sqrt{\frac{1805}{16}}$$

Simplify.

$$x+\frac{49}{4}=\frac{19\sqrt{5}}{4} x+\frac{49}{4}=-\frac{19\sqrt{5}}{4}$$

Subtract \frac{49}{4} from both sides of the equation.

$$x=\frac{19\sqrt{5}-49}{4} x=\frac{-19\sqrt{5}-49}{4}$$