Question

Pls help pls show working brainliest given to clearest answer

Answer 1

Answer:

$x_1=\frac{1+\sqrt{61}}{6}\approx1.47,x_2=\frac{1-\sqrt{61}}{6}\approx-1.14$

Step-by-step explanation:

To solve a non-factorable quadratic equation in the form of $ax^2+bx+c=0$, you need to use the quadratic formula where $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

$3x^2+x-5=0$

$x=\frac{-1\pm\sqrt{1^2-4(3)(-5)}}{2(3)}$

$x=\frac{-1\pm\sqrt{1+60}}{6}$

$x=\frac{-1\pm\sqrt{61}}{6}$

Because of the $\pm$ sign, quadratic equations always have two solutions, whether both are real, only one is real, or both are complex (in the form of $a+bi$ where $a$ is the real part and $b$ is the imaginary part where $i=\sqrt{-1}$). In this case, because the discriminant is positive (defined as $b^2-4ac$), then there are 2 real solutions.

The solutions are thus $x_1=\frac{1+\sqrt{61}}{6}\approx1.47,x_2=\frac{1-\sqrt{61}}{6}\approx-1.14$