Answer:
[tex]x_1=\frac{1+\sqrt{61}}{6}\approx1.47,x_2=\frac{1-\sqrt{61}}{6}\approx-1.14[/tex]
Step-by-step explanation:
To solve a non-factorable quadratic equation in the form of [tex]ax^2+bx+c=0[/tex], you need to use the quadratic formula where [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]:
[tex]3x^2+x-5=0[/tex]
[tex]x=\frac{-1\pm\sqrt{1^2-4(3)(-5)}}{2(3)}[/tex]
[tex]x=\frac{-1\pm\sqrt{1+60}}{6}[/tex]
[tex]x=\frac{-1\pm\sqrt{61}}{6}[/tex]
Because of the [tex]\pm[/tex] sign, quadratic equations always have two solutions, whether both are real, only one is real, or both are complex (in the form of [tex]a+bi[/tex] where [tex]a[/tex] is the real part and [tex]b[/tex] is the imaginary part where [tex]i=\sqrt{-1}[/tex]). In this case, because the discriminant is positive (defined as [tex]b^2-4ac[/tex]), then there are 2 real solutions.
The solutions are thus [tex]x_1=\frac{1+\sqrt{61}}{6}\approx1.47,x_2=\frac{1-\sqrt{61}}{6}\approx-1.14[/tex]