Given:
[tex]-16x^2-12x+10=0[/tex]
Find:
Value of x with the help of quadratic formula.
Sol:
Quadratic formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]
Use the formula :
[tex]\begin{gathered} -16x^2-12x+10=0 \\ \\ x=\frac{-(-12)\pm\sqrt{(-12)^2-4(-16)(10)}}{2(-16)} \\ \\ x=\frac{12\pm\sqrt{144+640}}{-32} \\ \\ x=\frac{12\pm\sqrt{784}}{-32} \\ \\ x=\frac{12\pm28}{-32} \\ \\ x=\frac{12+28}{-32};x=\frac{12-28}{-32} \\ \\ x=-1.25;x=0.5 \end{gathered}[/tex]
So the value of "x" is -1.25 and 0.5
