Solve 32x^2+250x-1925=0

$$32 { x }^{ 2 } +250x-1925=0$$

Solve for x

$$x = \frac{5 \sqrt{3089} - 125}{32} \approx 4.77793327$$

$$x=\frac{-5\sqrt{3089}-125}{32}\approx -12.59043327$$

Steps Using the Quadratic Formula

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.

$$32x^{2}+250x-1925=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 32 for a, 250 for b, and -1925 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-250±\sqrt{250^{2}-4\times 32\left(-1925\right)}}{2\times 32}$$

Square 250.

$$x=\frac{-250±\sqrt{62500-4\times 32\left(-1925\right)}}{2\times 32}$$

Multiply -4 times 32.

$$x=\frac{-250±\sqrt{62500-128\left(-1925\right)}}{2\times 32}$$

Multiply -128 times -1925.

$$x=\frac{-250±\sqrt{62500+246400}}{2\times 32}$$

Add 62500 to 246400.

$$x=\frac{-250±\sqrt{308900}}{2\times 32}$$

Take the square root of 308900.

$$x=\frac{-250±10\sqrt{3089}}{2\times 32}$$

Multiply 2 times 32.

$$x=\frac{-250±10\sqrt{3089}}{64}$$

Now solve the equation x=\frac{-250±10\sqrt{3089}}{64} when ± is plus. Add -250 to 10\sqrt{3089}.

$$x=\frac{10\sqrt{3089}-250}{64}$$

Divide -250+10\sqrt{3089} by 64.

$$x=\frac{5\sqrt{3089}-125}{32}$$

Now solve the equation x=\frac{-250±10\sqrt{3089}}{64} when ± is minus. Subtract 10\sqrt{3089} from -250.

$$x=\frac{-10\sqrt{3089}-250}{64}$$

Divide -250-10\sqrt{3089} by 64.

$$x=\frac{-5\sqrt{3089}-125}{32}$$

The equation is now solved.

$$x=\frac{5\sqrt{3089}-125}{32} x=\frac{-5\sqrt{3089}-125}{32}$$

Steps for Completing the Square

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.

$$32x^{2}+250x-1925=0$$

Add 1925 to both sides of the equation.

$$32x^{2}+250x-1925-\left(-1925\right)=-\left(-1925\right)$$

Subtracting -1925 from itself leaves 0.

$$32x^{2}+250x=-\left(-1925\right)$$

Subtract -1925 from 0.

$$32x^{2}+250x=1925$$

Divide both sides by 32.

$$\frac{32x^{2}+250x}{32}=\frac{1925}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x^{2}+\frac{250}{32}x=\frac{1925}{32}$$

Reduce the fraction \frac{250}{32} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{125}{16}x=\frac{1925}{32}$$

Divide \frac{125}{16}, the coefficient of the x term, by 2 to get \frac{125}{32}. Then add the square of \frac{125}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{125}{16}x+\left(\frac{125}{32}\right)^{2}=\frac{1925}{32}+\left(\frac{125}{32}\right)^{2}$$

Square \frac{125}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{125}{16}x+\frac{15625}{1024}=\frac{1925}{32}+\frac{15625}{1024}$$

Add \frac{1925}{32} to \frac{15625}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{125}{16}x+\frac{15625}{1024}=\frac{77225}{1024}$$

Factor x^{2}+\frac{125}{16}x+\frac{15625}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{125}{32}\right)^{2}=\frac{77225}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{125}{32}\right)^{2}}=\sqrt{\frac{77225}{1024}}$$

Simplify.

$$x+\frac{125}{32}=\frac{5\sqrt{3089}}{32} x+\frac{125}{32}=-\frac{5\sqrt{3089}}{32}$$

Subtract \frac{125}{32} from both sides of the equation.

$$x=\frac{5\sqrt{3089}-125}{32} x=\frac{-5\sqrt{3089}-125}{32}$$