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-x^{2}-3x-5=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(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9+4\left(-5\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-3\right)±\sqrt{9-20}}{2\left(-1\right)}
Multiply 4 times -5.
x=\frac{-\left(-3\right)±\sqrt{-11}}{2\left(-1\right)}
Add 9 to -20.
x=\frac{-\left(-3\right)±\sqrt{11}i}{2\left(-1\right)}
Take the square root of -11.
x=\frac{3±\sqrt{11}i}{2\left(-1\right)}
The opposite of -3 is 3.
x=\frac{3±\sqrt{11}i}{-2}
Multiply 2 times -1.
x=\frac{3+\sqrt{11}i}{-2}
Now solve the equation x=\frac{3±\sqrt{11}i}{-2} when ± is plus. Add 3 to i\sqrt{11}.
x=\frac{-\sqrt{11}i-3}{2}
Divide 3+i\sqrt{11} by -2.
x=\frac{-\sqrt{11}i+3}{-2}
Now solve the equation x=\frac{3±\sqrt{11}i}{-2} when ± is minus. Subtract i\sqrt{11} from 3.
x=\frac{-3+\sqrt{11}i}{2}
Divide 3-i\sqrt{11} by -2.
x=\frac{-\sqrt{11}i-3}{2} x=\frac{-3+\sqrt{11}i}{2}
The equation is now solved.
-x^{2}-3x-5=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}-3x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
-x^{2}-3x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
-x^{2}-3x=5
Subtract -5 from 0.
\frac{-x^{2}-3x}{-1}=\frac{5}{-1}
Divide both sides by -1.
x^{2}+\left(-\frac{3}{-1}\right)x=\frac{5}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+3x=\frac{5}{-1}
Divide -3 by -1.
x^{2}+3x=-5
Divide 5 by -1.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-5+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+3x+\frac{9}{4}=-5+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=-\frac{11}{4}
Add -5 to \frac{9}{4}.
\left(x+\frac{3}{2}\right)^{2}=-\frac{11}{4}
Factor x^{2}+3x+\frac{9}{4}. 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{3}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{\sqrt{11}i}{2} x+\frac{3}{2}=-\frac{\sqrt{11}i}{2}
Simplify.
x=\frac{-3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-3}{2}
Subtract \frac{3}{2} from both sides of the equation.
x ^ 2 +3x +5 = 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 = -3 rs = 5
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{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. 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{3}{2} - u) (-\frac{3}{2} + u) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
\frac{9}{4} - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-\frac{9}{4} = \frac{11}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = -\frac{11}{4} u = \pm\sqrt{-\frac{11}{4}} = \pm \frac{\sqrt{11}}{2}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{2} - \frac{\sqrt{11}}{2}i = -1.500 - 1.658i s = -\frac{3}{2} + \frac{\sqrt{11}}{2}i = -1.500 + 1.658i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.