Solve 4(x+3)^2=25(x-2)^2

$$4 ( x + 3 ) ^ { 2 } = 25 ( x - 2 ) ^ { 2 }$$

Solve for x

$$x = \frac{16}{3} = 5\frac{1}{3} \approx 5.333333333$$

$$x=\frac{4}{7}\approx 0.571428571$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$4\left(x^{2}+6x+9\right)=25\left(x-2\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=25\left(x-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$4x^{2}+24x+36=25\left(x^{2}-4x+4\right)$$

Use the distributive property to multiply 25 by x^{2}-4x+4.

$$4x^{2}+24x+36=25x^{2}-100x+100$$

Subtract 25x^{2} from both sides.

$$4x^{2}+24x+36-25x^{2}=-100x+100$$

Combine 4x^{2} and -25x^{2} to get -21x^{2}.

$$-21x^{2}+24x+36=-100x+100$$

Add 100x to both sides.

$$-21x^{2}+24x+36+100x=100$$

Combine 24x and 100x to get 124x.

$$-21x^{2}+124x+36=100$$

Subtract 100 from both sides.

$$-21x^{2}+124x+36-100=0$$

Subtract 100 from 36 to get -64.

$$-21x^{2}+124x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -21 for a, 124 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-124±\sqrt{124^{2}-4\left(-21\right)\left(-64\right)}}{2\left(-21\right)}$$

Square 124.

$$x=\frac{-124±\sqrt{15376-4\left(-21\right)\left(-64\right)}}{2\left(-21\right)}$$

Multiply -4 times -21.

$$x=\frac{-124±\sqrt{15376+84\left(-64\right)}}{2\left(-21\right)}$$

Multiply 84 times -64.

$$x=\frac{-124±\sqrt{15376-5376}}{2\left(-21\right)}$$

Add 15376 to -5376.

$$x=\frac{-124±\sqrt{10000}}{2\left(-21\right)}$$

Take the square root of 10000.

$$x=\frac{-124±100}{2\left(-21\right)}$$

Multiply 2 times -21.

$$x=\frac{-124±100}{-42}$$

Now solve the equation x=\frac{-124±100}{-42} when ± is plus. Add -124 to 100.

$$x=-\frac{24}{-42}$$

Reduce the fraction \frac{-24}{-42} to lowest terms by extracting and canceling out 6.

$$x=\frac{4}{7}$$

Now solve the equation x=\frac{-124±100}{-42} when ± is minus. Subtract 100 from -124.

$$x=-\frac{224}{-42}$$

Reduce the fraction \frac{-224}{-42} to lowest terms by extracting and canceling out 14.

$$x=\frac{16}{3}$$

The equation is now solved.

$$x=\frac{4}{7} x=\frac{16}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$4\left(x^{2}+6x+9\right)=25\left(x-2\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=25\left(x-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$4x^{2}+24x+36=25\left(x^{2}-4x+4\right)$$

Use the distributive property to multiply 25 by x^{2}-4x+4.

$$4x^{2}+24x+36=25x^{2}-100x+100$$

Subtract 25x^{2} from both sides.

$$4x^{2}+24x+36-25x^{2}=-100x+100$$

Combine 4x^{2} and -25x^{2} to get -21x^{2}.

$$-21x^{2}+24x+36=-100x+100$$

Add 100x to both sides.

$$-21x^{2}+24x+36+100x=100$$

Combine 24x and 100x to get 124x.

$$-21x^{2}+124x+36=100$$

Subtract 36 from both sides.

$$-21x^{2}+124x=100-36$$

Subtract 36 from 100 to get 64.

$$-21x^{2}+124x=64$$

Divide both sides by -21.

$$\frac{-21x^{2}+124x}{-21}=\frac{64}{-21}$$

Dividing by -21 undoes the multiplication by -21.

$$x^{2}+\frac{124}{-21}x=\frac{64}{-21}$$

Divide 124 by -21.

$$x^{2}-\frac{124}{21}x=\frac{64}{-21}$$

Divide 64 by -21.

$$x^{2}-\frac{124}{21}x=-\frac{64}{21}$$

Divide -\frac{124}{21}, the coefficient of the x term, by 2 to get -\frac{62}{21}. Then add the square of -\frac{62}{21} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{124}{21}x+\left(-\frac{62}{21}\right)^{2}=-\frac{64}{21}+\left(-\frac{62}{21}\right)^{2}$$

Square -\frac{62}{21} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{124}{21}x+\frac{3844}{441}=-\frac{64}{21}+\frac{3844}{441}$$

Add -\frac{64}{21} to \frac{3844}{441} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{124}{21}x+\frac{3844}{441}=\frac{2500}{441}$$

Factor x^{2}-\frac{124}{21}x+\frac{3844}{441}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{62}{21}\right)^{2}=\frac{2500}{441}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{62}{21}\right)^{2}}=\sqrt{\frac{2500}{441}}$$

Simplify.

$$x-\frac{62}{21}=\frac{50}{21} x-\frac{62}{21}=-\frac{50}{21}$$

Add \frac{62}{21} to both sides of the equation.

$$x=\frac{16}{3} x=\frac{4}{7}$$