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x^{2}+10x-15=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{-10±\sqrt{10^{2}-4\left(-15\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\left(-15\right)}}{2}
Square 10.
x=\frac{-10±\sqrt{100+60}}{2}
Multiply -4 times -15.
x=\frac{-10±\sqrt{160}}{2}
Add 100 to 60.
x=\frac{-10±4\sqrt{10}}{2}
Take the square root of 160.
x=\frac{4\sqrt{10}-10}{2}
Now solve the equation x=\frac{-10±4\sqrt{10}}{2} when ± is plus. Add -10 to 4\sqrt{10}.
x=2\sqrt{10}-5
Divide -10+4\sqrt{10} by 2.
x=\frac{-4\sqrt{10}-10}{2}
Now solve the equation x=\frac{-10±4\sqrt{10}}{2} when ± is minus. Subtract 4\sqrt{10} from -10.
x=-2\sqrt{10}-5
Divide -10-4\sqrt{10} by 2.
x=2\sqrt{10}-5 x=-2\sqrt{10}-5
The equation is now solved.
x^{2}+10x-15=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.
x^{2}+10x-15-\left(-15\right)=-\left(-15\right)
Add 15 to both sides of the equation.
x^{2}+10x=-\left(-15\right)
Subtracting -15 from itself leaves 0.
x^{2}+10x=15
Subtract -15 from 0.
x^{2}+10x+5^{2}=15+5^{2}
Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+10x+25=15+25
Square 5.
x^{2}+10x+25=40
Add 15 to 25.
\left(x+5\right)^{2}=40
Factor x^{2}+10x+25. 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+5\right)^{2}}=\sqrt{40}
Take the square root of both sides of the equation.
x+5=2\sqrt{10} x+5=-2\sqrt{10}
Simplify.
x=2\sqrt{10}-5 x=-2\sqrt{10}-5
Subtract 5 from both sides of the equation.
x ^ 2 +10x -15 = 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.
r + s = -10 rs = -15
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 = -5 - u s = -5 + u
Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. 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>
(-5 - u) (-5 + u) = -15
To solve for unknown quantity u, substitute these in the product equation rs = -15
25 - u^2 = -15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -15-25 = -40
Simplify the expression by subtracting 25 on both sides
u^2 = 40 u = \pm\sqrt{40} = \pm \sqrt{40}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-5 - \sqrt{40} = -11.325 s = -5 + \sqrt{40} = 1.325
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.