\(\sf~x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Our equation is in the form of \(\sf~ax^2+bx+c\)
So in this case:
\(\sf~a=1\)
\(\sf~b=7\)
\(\sf~c=8\)
\(\sf~x=\dfrac{-7\pm\sqrt{7^2-4(1)(8)}}{2(1)}\)
Simplify exponent and the denominator:
\(\sf~x=\dfrac{-7\pm\sqrt{49-4(1)(8)}}{2}\)
Multiply:
\(\sf~x=\dfrac{-7\pm\sqrt{49-32}}{2}\)
Subtract:
\(\sf~x=\dfrac{-7\pm\sqrt{17}}{2}\)
Simplify the square root:
\(\sf~x\approx\dfrac{-7\pm4.12310563}{2}\)
Now this breaks into two equations.
\(\sf~x\approx\dfrac{-7+4.12310563}{2}\)
and
\(\sf~x\approx\dfrac{-7-4.12310563}{2}\)
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\(\sf~x\approx\dfrac{-7+4.12310563}{2}\)
Add:
\(\sf~x\approx\dfrac{-2.87689437}{2}\)
Divide:
\(\sf~x\approx-1.43844719\)
Round to the nearest tenth:
\(\sf~x\approx\boxed{\sf-1.44}\)
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\(\sf~x\approx\dfrac{-7-4.12310563}{2}\)
Subtract:
\(\sf~x\approx\dfrac{-11.1231056}{2}\)
Divide:
\(\sf~x\approx-5.5615528\)
Round to the nearest tenth:
\(\sf~x\approx\boxed{\sf-5.56}\)
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Your answers were correct.
