Question

7x^2+9x-1=0 solve the quadratic equation ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$x=\dfrac{-9+\sqrt{109}}{14},\:x=\dfrac{-9-\sqrt{109}}{14}\quad\left(\mathrm{Decimal:\:}&x=0.10288\dots,x=-1.38859\dots\right)$

Step-by-step explanation:

$7x^2+9x-1=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

• $\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$ $x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=7,\:b=9,\:c=-1:\quad x_{1,\:2}=\dfrac{-9\pm \sqrt{9^2-4\cdot \:7\left(-1\right)}}{2\cdot \:7}$

$x=\dfrac{-9+\sqrt{9^2-4\cdot \:7\left(-1\right)}}{2\cdot \:7}$

$\dfrac{-9+\sqrt{9^2-4\cdot \:7\left(-1\right)}}{2\cdot \:7}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\dfrac{-9+\sqrt{9^2+4\cdot \:7\cdot \:1}}{2\cdot \:7}$

$\sqrt{9^2+4\cdot \:7\cdot \:1}$

$9^2=81$

$=\sqrt{81+4\cdot \:7\cdot \:1}$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:7\cdot \:1=28$

$=\sqrt{81+28}$

$\mathrm{Add\:the\:numbers:}\:81+28=109$

$=\sqrt{109}$

$=\dfrac{-9+\sqrt{109}}{2\cdot \:7}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:7=14$

$=\dfrac{-9+\sqrt{109}}{14}$

$x=\dfrac{-9-\sqrt{9^2-4\cdot \:7\left(-1\right)}}{2\cdot \:7}$

$\dfrac{-9-\sqrt{9^2-4\cdot \:7\left(-1\right)}}{2\cdot \:7}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\dfrac{-9-\sqrt{9^2+4\cdot \:7\cdot \:1}}{2\cdot \:7}$

$\sqrt{9^2+4\cdot \:7\cdot \:1}=\sqrt{109}$

$=\dfrac{-9-\sqrt{109}}{2\cdot \:7}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:7=14$

$=\dfrac{-9-\sqrt{109}}{14}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\dfrac{-9+\sqrt{109}}{14},\:x=\dfrac{-9-\sqrt{109}}{14}$