Respuesta :
You can rearrange the equation as [tex] x^2+13x+4 = 0 [/tex]
To solve quadratic equations like [tex] ax^2+bx+c = 0 [/tex] we use the quadratic formula:
[tex] x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
In your case, [tex] a = 1, b = 13, c = 4 [/tex], so the formula becomes
[tex] x_{1,2} = \cfrac{-13\pm\sqrt{169-16}}{2} =\cfrac{-13\pm\sqrt{153}}{2} [/tex]
So, the two solutions are
[tex] x = \cfrac{-13+\sqrt{153}}{2} \text{ or } x = \cfrac{-13-\sqrt{153}}{2} [/tex]
The solutions of the equation are -12.68 and -0.31.
What is a quadratic equation?
Quadratic equations have the form ax2 + bx + c = 0 and are second-degree algebraic expressions.
The equation given in the question is x^2 + 13x + 4 = 0
D = b^2 - 4ac
D = (13)^2 - 4*1*4
D = 153
Roots of the equation are:
x1 = (-b - √D)/2a
x1 = (-13 - √153)/ 2*(1)
x1 = (-25.36)/2
x1 = -12.68
x2 =(-b + √D)/2a
x2 = (-13 + √153)/2*(1)
x2 = -0.63/2
x2 = -0.31
Hence the roots of the quadratic equation are -0.31 and -12.68.
Learn more about quadratic equations on:
https://brainly.com/question/1214333
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