Solve -0.7x^2+105x-1250=0

$$- 0.7 x ^ { 2 } + 105 x - 1250 = 0$$

Solve for x

$$x=\frac{25\sqrt{301}}{7}+75\approx 136.961969903$$

$$x=-\frac{25\sqrt{301}}{7}+75\approx 13.038030097$$

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.

$$-0.7x^{2}+105x-1250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.7 for a, 105 for b, and -1250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-105±\sqrt{105^{2}-4\left(-0.7\right)\left(-1250\right)}}{2\left(-0.7\right)}$$

Square 105.

$$x=\frac{-105±\sqrt{11025-4\left(-0.7\right)\left(-1250\right)}}{2\left(-0.7\right)}$$

Multiply -4 times -0.7.

$$x=\frac{-105±\sqrt{11025+2.8\left(-1250\right)}}{2\left(-0.7\right)}$$

Multiply 2.8 times -1250.

$$x=\frac{-105±\sqrt{11025-3500}}{2\left(-0.7\right)}$$

Add 11025 to -3500.

$$x=\frac{-105±\sqrt{7525}}{2\left(-0.7\right)}$$

Take the square root of 7525.

$$x=\frac{-105±5\sqrt{301}}{2\left(-0.7\right)}$$

Multiply 2 times -0.7.

$$x=\frac{-105±5\sqrt{301}}{-1.4}$$

Now solve the equation x=\frac{-105±5\sqrt{301}}{-1.4} when ± is plus. Add -105 to 5\sqrt{301}.

$$x=\frac{5\sqrt{301}-105}{-1.4}$$

Divide -105+5\sqrt{301} by -1.4 by multiplying -105+5\sqrt{301} by the reciprocal of -1.4.

$$x=-\frac{25\sqrt{301}}{7}+75$$

Now solve the equation x=\frac{-105±5\sqrt{301}}{-1.4} when ± is minus. Subtract 5\sqrt{301} from -105.

$$x=\frac{-5\sqrt{301}-105}{-1.4}$$

Divide -105-5\sqrt{301} by -1.4 by multiplying -105-5\sqrt{301} by the reciprocal of -1.4.

$$x=\frac{25\sqrt{301}}{7}+75$$

The equation is now solved.

$$x=-\frac{25\sqrt{301}}{7}+75 x=\frac{25\sqrt{301}}{7}+75$$

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.

$$-0.7x^{2}+105x-1250=0$$

Add 1250 to both sides of the equation.

$$-0.7x^{2}+105x-1250-\left(-1250\right)=-\left(-1250\right)$$

Subtracting -1250 from itself leaves 0.

$$-0.7x^{2}+105x=-\left(-1250\right)$$

Subtract -1250 from 0.

$$-0.7x^{2}+105x=1250$$

Divide both sides of the equation by -0.7, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-0.7x^{2}+105x}{-0.7}=\frac{1250}{-0.7}$$

Dividing by -0.7 undoes the multiplication by -0.7.

$$x^{2}+\frac{105}{-0.7}x=\frac{1250}{-0.7}$$

Divide 105 by -0.7 by multiplying 105 by the reciprocal of -0.7.

$$x^{2}-150x=\frac{1250}{-0.7}$$

Divide 1250 by -0.7 by multiplying 1250 by the reciprocal of -0.7.

$$x^{2}-150x=-\frac{12500}{7}$$

Divide -150, the coefficient of the x term, by 2 to get -75. Then add the square of -75 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-150x+\left(-75\right)^{2}=-\frac{12500}{7}+\left(-75\right)^{2}$$

Square -75.

$$x^{2}-150x+5625=-\frac{12500}{7}+5625$$

Add -\frac{12500}{7} to 5625.

$$x^{2}-150x+5625=\frac{26875}{7}$$

Factor x^{2}-150x+5625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-75\right)^{2}=\frac{26875}{7}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-75\right)^{2}}=\sqrt{\frac{26875}{7}}$$

Simplify.

$$x-75=\frac{25\sqrt{301}}{7} x-75=-\frac{25\sqrt{301}}{7}$$

Add 75 to both sides of the equation.

$$x=\frac{25\sqrt{301}}{7}+75 x=-\frac{25\sqrt{301}}{7}+75$$