Solve (5x-4)*(5x+4)/4=(5x-7)^2-9/2

$$\frac { ( 5 x - 4 ) \cdot ( 5 x + 4 ) } { 4 } = \frac { ( 5 x - 7 ) ^ { 2 } - 9 } { 2 }$$

Solve for x

$$x = \frac{24}{5} = 4\frac{4}{5} = 4.8$$

$$x=\frac{4}{5}=0.8$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$\left(5x-4\right)\left(5x+4\right)=2\left(\left(5x-7\right)^{2}-9\right)$$

Consider \left(5x-4\right)\left(5x+4\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 4.

$$\left(5x\right)^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-7\right)^{2}.

$$25x^{2}-16=2\left(25x^{2}-70x+49-9\right)$$

Subtract 9 from 49 to get 40.

$$25x^{2}-16=2\left(25x^{2}-70x+40\right)$$

Use the distributive property to multiply 2 by 25x^{2}-70x+40.

$$25x^{2}-16=50x^{2}-140x+80$$

Subtract 50x^{2} from both sides.

$$25x^{2}-16-50x^{2}=-140x+80$$

Combine 25x^{2} and -50x^{2} to get -25x^{2}.

$$-25x^{2}-16=-140x+80$$

Add 140x to both sides.

$$-25x^{2}-16+140x=80$$

Subtract 80 from both sides.

$$-25x^{2}-16+140x-80=0$$

Subtract 80 from -16 to get -96.

$$-25x^{2}-96+140x=0$$

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.

$$-25x^{2}+140x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 140 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-140±\sqrt{140^{2}-4\left(-25\right)\left(-96\right)}}{2\left(-25\right)}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\left(-25\right)\left(-96\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-140±\sqrt{19600+100\left(-96\right)}}{2\left(-25\right)}$$

Multiply 100 times -96.

$$x=\frac{-140±\sqrt{19600-9600}}{2\left(-25\right)}$$

Add 19600 to -9600.

$$x=\frac{-140±\sqrt{10000}}{2\left(-25\right)}$$

Take the square root of 10000.

$$x=\frac{-140±100}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-140±100}{-50}$$

Now solve the equation x=\frac{-140±100}{-50} when ± is plus. Add -140 to 100.

$$x=-\frac{40}{-50}$$

Reduce the fraction \frac{-40}{-50} to lowest terms by extracting and canceling out 10.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{-140±100}{-50} when ± is minus. Subtract 100 from -140.

$$x=-\frac{240}{-50}$$

Reduce the fraction \frac{-240}{-50} to lowest terms by extracting and canceling out 10.

$$x=\frac{24}{5}$$

The equation is now solved.

$$x=\frac{4}{5} x=\frac{24}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$\left(5x-4\right)\left(5x+4\right)=2\left(\left(5x-7\right)^{2}-9\right)$$

Consider \left(5x-4\right)\left(5x+4\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 4.

$$\left(5x\right)^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-16=2\left(\left(5x-7\right)^{2}-9\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-7\right)^{2}.

$$25x^{2}-16=2\left(25x^{2}-70x+49-9\right)$$

Subtract 9 from 49 to get 40.

$$25x^{2}-16=2\left(25x^{2}-70x+40\right)$$

Use the distributive property to multiply 2 by 25x^{2}-70x+40.

$$25x^{2}-16=50x^{2}-140x+80$$

Subtract 50x^{2} from both sides.

$$25x^{2}-16-50x^{2}=-140x+80$$

Combine 25x^{2} and -50x^{2} to get -25x^{2}.

$$-25x^{2}-16=-140x+80$$

Add 140x to both sides.

$$-25x^{2}-16+140x=80$$

Add 16 to both sides.

$$-25x^{2}+140x=80+16$$

Add 80 and 16 to get 96.

$$-25x^{2}+140x=96$$

Divide both sides by -25.

$$\frac{-25x^{2}+140x}{-25}=\frac{96}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$x^{2}+\frac{140}{-25}x=\frac{96}{-25}$$

Reduce the fraction \frac{140}{-25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{28}{5}x=\frac{96}{-25}$$

Divide 96 by -25.

$$x^{2}-\frac{28}{5}x=-\frac{96}{25}$$

Divide -\frac{28}{5}, the coefficient of the x term, by 2 to get -\frac{14}{5}. Then add the square of -\frac{14}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{28}{5}x+\left(-\frac{14}{5}\right)^{2}=-\frac{96}{25}+\left(-\frac{14}{5}\right)^{2}$$

Square -\frac{14}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{28}{5}x+\frac{196}{25}=\frac{-96+196}{25}$$

Add -\frac{96}{25} to \frac{196}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{28}{5}x+\frac{196}{25}=4$$

Factor x^{2}-\frac{28}{5}x+\frac{196}{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}.

$$\left(x-\frac{14}{5}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{14}{5}\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-\frac{14}{5}=2 x-\frac{14}{5}=-2$$

Add \frac{14}{5} to both sides of the equation.

$$x=\frac{24}{5} x=\frac{4}{5}$$