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4x^{2}+6x+10=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{-6±\sqrt{6^{2}-4\times 4\times 10}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 6 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 4\times 10}}{2\times 4}
Square 6.
x=\frac{-6±\sqrt{36-16\times 10}}{2\times 4}
Multiply -4 times 4.
x=\frac{-6±\sqrt{36-160}}{2\times 4}
Multiply -16 times 10.
x=\frac{-6±\sqrt{-124}}{2\times 4}
Add 36 to -160.
x=\frac{-6±2\sqrt{31}i}{2\times 4}
Take the square root of -124.
x=\frac{-6±2\sqrt{31}i}{8}
Multiply 2 times 4.
x=\frac{-6+2\sqrt{31}i}{8}
Now solve the equation x=\frac{-6±2\sqrt{31}i}{8} when ± is plus. Add -6 to 2i\sqrt{31}.
x=\frac{-3+\sqrt{31}i}{4}
Divide -6+2i\sqrt{31} by 8.
x=\frac{-2\sqrt{31}i-6}{8}
Now solve the equation x=\frac{-6±2\sqrt{31}i}{8} when ± is minus. Subtract 2i\sqrt{31} from -6.
x=\frac{-\sqrt{31}i-3}{4}
Divide -6-2i\sqrt{31} by 8.
x=\frac{-3+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-3}{4}
The equation is now solved.
4x^{2}+6x+10=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.
4x^{2}+6x+10-10=-10
Subtract 10 from both sides of the equation.
4x^{2}+6x=-10
Subtracting 10 from itself leaves 0.
\frac{4x^{2}+6x}{4}=-\frac{10}{4}
Divide both sides by 4.
x^{2}+\frac{6}{4}x=-\frac{10}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{3}{2}x=-\frac{10}{4}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{3}{2}x=-\frac{5}{2}
Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{3}{4}\right)^{2}
Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{5}{2}+\frac{9}{16}
Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{31}{16}
Add -\frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{4}\right)^{2}=-\frac{31}{16}
Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. 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}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}
Take the square root of both sides of the equation.
x+\frac{3}{4}=\frac{\sqrt{31}i}{4} x+\frac{3}{4}=-\frac{\sqrt{31}i}{4}
Simplify.
x=\frac{-3+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-3}{4}
Subtract \frac{3}{4} from both sides of the equation.
x ^ 2 +\frac{3}{2}x +\frac{5}{2} = 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 4
r + s = -\frac{3}{2} rs = \frac{5}{2}
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}{4} - u s = -\frac{3}{4} + u
Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}. 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}{4} - u) (-\frac{3}{4} + u) = \frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}
\frac{9}{16} - u^2 = \frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{2}-\frac{9}{16} = \frac{31}{16}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = -\frac{31}{16} u = \pm\sqrt{-\frac{31}{16}} = \pm \frac{\sqrt{31}}{4}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}{4} - \frac{\sqrt{31}}{4}i = -0.750 - 1.392i s = -\frac{3}{4} + \frac{\sqrt{31}}{4}i = -0.750 + 1.392i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.