Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

7x^{2}-10x-1000=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.
x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 7\left(-1000\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -10 for b, and -1000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 7\left(-1000\right)}}{2\times 7}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-28\left(-1000\right)}}{2\times 7}
Multiply -4 times 7.
x=\frac{-\left(-10\right)±\sqrt{100+28000}}{2\times 7}
Multiply -28 times -1000.
x=\frac{-\left(-10\right)±\sqrt{28100}}{2\times 7}
Add 100 to 28000.
x=\frac{-\left(-10\right)±10\sqrt{281}}{2\times 7}
Take the square root of 28100.
x=\frac{10±10\sqrt{281}}{2\times 7}
The opposite of -10 is 10.
x=\frac{10±10\sqrt{281}}{14}
Multiply 2 times 7.
x=\frac{10\sqrt{281}+10}{14}
Now solve the equation x=\frac{10±10\sqrt{281}}{14} when ± is plus. Add 10 to 10\sqrt{281}.
x=\frac{5\sqrt{281}+5}{7}
Divide 10+10\sqrt{281} by 14.
x=\frac{10-10\sqrt{281}}{14}
Now solve the equation x=\frac{10±10\sqrt{281}}{14} when ± is minus. Subtract 10\sqrt{281} from 10.
x=\frac{5-5\sqrt{281}}{7}
Divide 10-10\sqrt{281} by 14.
x=\frac{5\sqrt{281}+5}{7} x=\frac{5-5\sqrt{281}}{7}
The equation is now solved.
7x^{2}-10x-1000=0
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.
7x^{2}-10x-1000-\left(-1000\right)=-\left(-1000\right)
Add 1000 to both sides of the equation.
7x^{2}-10x=-\left(-1000\right)
Subtracting -1000 from itself leaves 0.
7x^{2}-10x=1000
Subtract -1000 from 0.
\frac{7x^{2}-10x}{7}=\frac{1000}{7}
Divide both sides by 7.
x^{2}-\frac{10}{7}x=\frac{1000}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}-\frac{10}{7}x+\left(-\frac{5}{7}\right)^{2}=\frac{1000}{7}+\left(-\frac{5}{7}\right)^{2}
Divide -\frac{10}{7}, the coefficient of the x term, by 2 to get -\frac{5}{7}. Then add the square of -\frac{5}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{7}x+\frac{25}{49}=\frac{1000}{7}+\frac{25}{49}
Square -\frac{5}{7} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{7}x+\frac{25}{49}=\frac{7025}{49}
Add \frac{1000}{7} to \frac{25}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{7}\right)^{2}=\frac{7025}{49}
Factor x^{2}-\frac{10}{7}x+\frac{25}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{5}{7}\right)^{2}}=\sqrt{\frac{7025}{49}}
Take the square root of both sides of the equation.
x-\frac{5}{7}=\frac{5\sqrt{281}}{7} x-\frac{5}{7}=-\frac{5\sqrt{281}}{7}
Simplify.
x=\frac{5\sqrt{281}+5}{7} x=\frac{5-5\sqrt{281}}{7}
Add \frac{5}{7} to both sides of the equation.
x ^ 2 -\frac{10}{7}x -\frac{1000}{7} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 7
r + s = \frac{10}{7} rs = -\frac{1000}{7}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{5}{7} - u s = \frac{5}{7} + u
Two numbers r and s sum up to \frac{10}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{7} = \frac{5}{7}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{5}{7} - u) (\frac{5}{7} + u) = -\frac{1000}{7}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1000}{7}
\frac{25}{49} - u^2 = -\frac{1000}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1000}{7}-\frac{25}{49} = -\frac{7025}{49}
Simplify the expression by subtracting \frac{25}{49} on both sides
u^2 = \frac{7025}{49} u = \pm\sqrt{\frac{7025}{49}} = \pm \frac{\sqrt{7025}}{7}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{7} - \frac{\sqrt{7025}}{7} = -11.259 s = \frac{5}{7} + \frac{\sqrt{7025}}{7} = 12.688
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.