Sure! Let's solve the quadratic equation given:
[tex]\[ z^2 + 10z - 24 = 0 \][/tex]
The general form of a quadratic equation is:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = -24 \)[/tex].
To find the solutions for [tex]\( z \)[/tex], we can use the quadratic formula:
[tex]\[ z = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
First, let's calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant:
[tex]\[ \Delta = 10^2 - 4(1)(-24) \][/tex]
[tex]\[ \Delta = 100 + 96 \][/tex]
[tex]\[ \Delta = 196 \][/tex]
Now we substitute the values back into the quadratic formula:
[tex]\[ z = \frac{{-10 \pm \sqrt{196}}}{2(1)} \][/tex]
Since [tex]\( \sqrt{196} = 14 \)[/tex], this becomes:
[tex]\[ z = \frac{{-10 \pm 14}}{2} \][/tex]
This gives us two solutions:
[tex]\[ z = \frac{{-10 + 14}}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ z = \frac{{-10 - 14}}{2} = \frac{-24}{2} = -12 \][/tex]
Therefore, one of the solutions to the quadratic equation [tex]\( z^2 + 10z - 24 = 0 \)[/tex] is:
[tex]\[ z = -12 \][/tex]
