Solve left(4x+5right)left(4x-5right)=-10x+34

$$\left( 4x+5 \right) \left( 4x-5 \right) = -10x+34$$

Solve for x

$$x = \frac{\sqrt{969} - 5}{16} \approx 1.633047802$$

$$x=\frac{-\sqrt{969}-5}{16}\approx -2.258047802$$

Steps Using the Quadratic Formula

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

$$\left(4x\right)^{2}-25=-10x+34$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-25=-10x+34$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-25=-10x+34$$

Add 10x to both sides.

$$16x^{2}-25+10x=34$$

Subtract 34 from both sides.

$$16x^{2}-25+10x-34=0$$

Subtract 34 from -25 to get -59.

$$16x^{2}-59+10x=0$$

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.

$$16x^{2}+10x-59=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 10 for b, and -59 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 16\left(-59\right)}}{2\times 16}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 16\left(-59\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-10±\sqrt{100-64\left(-59\right)}}{2\times 16}$$

Multiply -64 times -59.

$$x=\frac{-10±\sqrt{100+3776}}{2\times 16}$$

Add 100 to 3776.

$$x=\frac{-10±\sqrt{3876}}{2\times 16}$$

Take the square root of 3876.

$$x=\frac{-10±2\sqrt{969}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-10±2\sqrt{969}}{32}$$

Now solve the equation x=\frac{-10±2\sqrt{969}}{32} when ± is plus. Add -10 to 2\sqrt{969}.

$$x=\frac{2\sqrt{969}-10}{32}$$

Divide -10+2\sqrt{969} by 32.

$$x=\frac{\sqrt{969}-5}{16}$$

Now solve the equation x=\frac{-10±2\sqrt{969}}{32} when ± is minus. Subtract 2\sqrt{969} from -10.

$$x=\frac{-2\sqrt{969}-10}{32}$$

Divide -10-2\sqrt{969} by 32.

$$x=\frac{-\sqrt{969}-5}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{969}-5}{16} x=\frac{-\sqrt{969}-5}{16}$$

Steps for Completing the Square

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

$$\left(4x\right)^{2}-25=-10x+34$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-25=-10x+34$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-25=-10x+34$$

Add 10x to both sides.

$$16x^{2}-25+10x=34$$

Add 25 to both sides.

$$16x^{2}+10x=34+25$$

Add 34 and 25 to get 59.

$$16x^{2}+10x=59$$

Divide both sides by 16.

$$\frac{16x^{2}+10x}{16}=\frac{59}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{10}{16}x=\frac{59}{16}$$

Reduce the fraction \frac{10}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{8}x=\frac{59}{16}$$

Divide \frac{5}{8}, the coefficient of the x term, by 2 to get \frac{5}{16}. Then add the square of \frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{8}x+\left(\frac{5}{16}\right)^{2}=\frac{59}{16}+\left(\frac{5}{16}\right)^{2}$$

Square \frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=\frac{59}{16}+\frac{25}{256}$$

Add \frac{59}{16} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=\frac{969}{256}$$

Factor x^{2}+\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{16}\right)^{2}=\frac{969}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{16}\right)^{2}}=\sqrt{\frac{969}{256}}$$

Simplify.

$$x+\frac{5}{16}=\frac{\sqrt{969}}{16} x+\frac{5}{16}=-\frac{\sqrt{969}}{16}$$

Subtract \frac{5}{16} from both sides of the equation.

$$x=\frac{\sqrt{969}-5}{16} x=\frac{-\sqrt{969}-5}{16}$$