To find the zeros of the quadratic function [tex]\( f(x) = 6x^2 + 12x - 7 \)[/tex], we need to use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 6 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -7 \)[/tex].
Next, we find the two solutions using the quadratic formula: [tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex] [tex]\[ x = \frac{-12 \pm \sqrt{312}}{2 \cdot 6} \][/tex] [tex]\[ x = \frac{-12 \pm \sqrt{312}}{12} \][/tex]
We simplify the square root of 312: [tex]\[ \sqrt{312} = \sqrt{4 \cdot 78} = 2\sqrt{78} \][/tex]
So, the solutions become: [tex]\[ x = \frac{-12 \pm 2\sqrt{78}}{12} \][/tex] [tex]\[ x = \frac{-12}{12} \pm \frac{2\sqrt{78}}{12} \][/tex] [tex]\[ x = -1 \pm \frac{\sqrt{78}}{6} \][/tex]
Therefore, the zeros of the quadratic function are: [tex]\[ x = -1 - \frac{\sqrt{78}}{6} \][/tex] [tex]\[ x = -1 + \frac{\sqrt{78}}{6} \][/tex]
If we compare these results with the given options, simplifying further: [tex]\[ \sqrt{78} \approx 8.83176 \][/tex]
