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x^{2}+6x+34=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 34}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 34}}{2}
Square 6.
x=\frac{-6±\sqrt{36-136}}{2}
Multiply -4 times 34.
x=\frac{-6±\sqrt{-100}}{2}
Add 36 to -136.
x=\frac{-6±10i}{2}
Take the square root of -100.
x=\frac{-6+10i}{2}
Now solve the equation x=\frac{-6±10i}{2} when ± is plus. Add -6 to 10i.
x=-3+5i
Divide -6+10i by 2.
x=\frac{-6-10i}{2}
Now solve the equation x=\frac{-6±10i}{2} when ± is minus. Subtract 10i from -6.
x=-3-5i
Divide -6-10i by 2.
x=-3+5i x=-3-5i
The equation is now solved.
x^{2}+6x+34=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}+6x+34-34=-34
Subtract 34 from both sides of the equation.
x^{2}+6x=-34
Subtracting 34 from itself leaves 0.
x^{2}+6x+3^{2}=-34+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+6x+9=-34+9
Square 3.
x^{2}+6x+9=-25
Add -34 to 9.
\left(x+3\right)^{2}=-25
Factor x^{2}+6x+9. 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+3\right)^{2}}=\sqrt{-25}
Take the square root of both sides of the equation.
x+3=5i x+3=-5i
Simplify.
x=-3+5i x=-3-5i
Subtract 3 from both sides of the equation.
x ^ 2 +6x +34 = 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 = -6 rs = 34
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 = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>
(-3 - u) (-3 + u) = 34
To solve for unknown quantity u, substitute these in the product equation rs = 34
9 - u^2 = 34
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 34-9 = 25
Simplify the expression by subtracting 9 on both sides
u^2 = -25 u = \pm\sqrt{-25} = \pm 5i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - 5i s = -3 + 5i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.