Question

1. solve for the values of x by factoring.

$5x^2 + 4x - 3 = 0$
$\bigcirc$ $x = \frac{-2 + 2\sqrt{19}}{5}; x = \frac{-2 - 2\sqrt{19}}{5}$
$\bigcirc$ $x = -3; x = 5$
$\bigcirc$ $x = 3; x = 5$
$\bigcirc$ $x = \frac{-2 + \sqrt{19}}{5}; x = \frac{-2 - \sqrt{19}}{5}$

Explanation:

Step1: Identify the quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the solutions are given by \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 5 \), \( b = 4 \), and \( c=-3 \).

Step2: Calculate the discriminant

The discriminant \( D=b^{2}-4ac=(4)^{2}-4\times5\times(-3)=16 + 60 = 76 \).

Step3: Simplify the square root of the discriminant

\( \sqrt{76}=\sqrt{4\times19}=2\sqrt{19} \), but when we plug into the quadratic formula, we have \( x=\frac{-4\pm2\sqrt{19}}{2\times5} \).

Step4: Simplify the fraction

Divide numerator and denominator by 2: \( x=\frac{-2\pm\sqrt{19}}{5} \), which gives \( x=\frac{-2+\sqrt{19}}{5} \) and \( x=\frac{-2-\sqrt{19}}{5} \).

Answer:

\( x=\frac{-2+\sqrt{19}}{5};x=\frac{-2-\sqrt{19}}{5} \)