Solve 1.9x^2+280x=-8100

$$1.9 x ^ { 2 } + 280 x = - 8100$$

Solve for x

$$x=\frac{10\sqrt{4210}-1400}{19}\approx -39.534466953$$

$$x=\frac{-10\sqrt{4210}-1400}{19}\approx -107.8339541$$

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.

$$1.9x^{2}+280x=-8100$$

Add 8100 to both sides of the equation.

$$1.9x^{2}+280x-\left(-8100\right)=-8100-\left(-8100\right)$$

Subtracting -8100 from itself leaves 0.

$$1.9x^{2}+280x-\left(-8100\right)=0$$

Subtract -8100 from 0.

$$1.9x^{2}+280x+8100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1.9 for a, 280 for b, and 8100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-280±\sqrt{280^{2}-4\times 1.9\times 8100}}{2\times 1.9}$$

Square 280.

$$x=\frac{-280±\sqrt{78400-4\times 1.9\times 8100}}{2\times 1.9}$$

Multiply -4 times 1.9.

$$x=\frac{-280±\sqrt{78400-7.6\times 8100}}{2\times 1.9}$$

Multiply -7.6 times 8100.

$$x=\frac{-280±\sqrt{78400-61560}}{2\times 1.9}$$

Add 78400 to -61560.

$$x=\frac{-280±\sqrt{16840}}{2\times 1.9}$$

Take the square root of 16840.

$$x=\frac{-280±2\sqrt{4210}}{2\times 1.9}$$

Multiply 2 times 1.9.

$$x=\frac{-280±2\sqrt{4210}}{3.8}$$

Now solve the equation x=\frac{-280±2\sqrt{4210}}{3.8} when ± is plus. Add -280 to 2\sqrt{4210}.

$$x=\frac{2\sqrt{4210}-280}{3.8}$$

Divide -280+2\sqrt{4210} by 3.8 by multiplying -280+2\sqrt{4210} by the reciprocal of 3.8.

$$x=\frac{10\sqrt{4210}-1400}{19}$$

Now solve the equation x=\frac{-280±2\sqrt{4210}}{3.8} when ± is minus. Subtract 2\sqrt{4210} from -280.

$$x=\frac{-2\sqrt{4210}-280}{3.8}$$

Divide -280-2\sqrt{4210} by 3.8 by multiplying -280-2\sqrt{4210} by the reciprocal of 3.8.

$$x=\frac{-10\sqrt{4210}-1400}{19}$$

The equation is now solved.

$$x=\frac{10\sqrt{4210}-1400}{19} x=\frac{-10\sqrt{4210}-1400}{19}$$

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.

$$1.9x^{2}+280x=-8100$$

Divide both sides of the equation by 1.9, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{1.9x^{2}+280x}{1.9}=-\frac{8100}{1.9}$$

Dividing by 1.9 undoes the multiplication by 1.9.

$$x^{2}+\frac{280}{1.9}x=-\frac{8100}{1.9}$$

Divide 280 by 1.9 by multiplying 280 by the reciprocal of 1.9.

$$x^{2}+\frac{2800}{19}x=-\frac{8100}{1.9}$$

Divide -8100 by 1.9 by multiplying -8100 by the reciprocal of 1.9.

$$x^{2}+\frac{2800}{19}x=-\frac{81000}{19}$$

Divide \frac{2800}{19}, the coefficient of the x term, by 2 to get \frac{1400}{19}. Then add the square of \frac{1400}{19} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2800}{19}x+\frac{1400}{19}^{2}=-\frac{81000}{19}+\frac{1400}{19}^{2}$$

Square \frac{1400}{19} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2800}{19}x+\frac{1960000}{361}=-\frac{81000}{19}+\frac{1960000}{361}$$

Add -\frac{81000}{19} to \frac{1960000}{361} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2800}{19}x+\frac{1960000}{361}=\frac{421000}{361}$$

Factor x^{2}+\frac{2800}{19}x+\frac{1960000}{361}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1400}{19}\right)^{2}=\frac{421000}{361}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1400}{19}\right)^{2}}=\sqrt{\frac{421000}{361}}$$

Simplify.

$$x+\frac{1400}{19}=\frac{10\sqrt{4210}}{19} x+\frac{1400}{19}=-\frac{10\sqrt{4210}}{19}$$

Subtract \frac{1400}{19} from both sides of the equation.

$$x=\frac{10\sqrt{4210}-1400}{19} x=\frac{-10\sqrt{4210}-1400}{19}$$