Given the equation:
[tex]-2x^2-4x=-5[/tex]STEP 1 of 2:
To solve the equation using the formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]We need to find the values of a, b, and c.
To do that, the equation must be in the form:
[tex]ax^2+bx+c=0[/tex]Let's rearrange the terms of the given equation:
[tex]-2x^2-4x+5=0[/tex]Now we can identify the values
a = -2, b = -4, c = 5
STEP 2 of 2: Calculate the value of the discriminant:
[tex]\begin{gathered} d=b^2-4ac \\ d=(-4)^2-4\cdot(-2)\cdot5=16+40=56 \end{gathered}[/tex]Since the discriminant is positive, the equation has two real solutions. Using the formula:
[tex]x=\frac{-(-4)\pm\sqrt[]{56}}{2\cdot(-2)}=\frac{4+\sqrt[]{56}}{-4}=\frac{4\pm7.483}{-4}[/tex]We have two solutions:
x = -2.87
x = 0.87
If we wanted to express the solutions in radical form, then we must simplify the expression:
[tex]x=\frac{4+\sqrt[]{56}}{-4}[/tex]Since 56 = 4 x 14 :
[tex]\begin{gathered} x=\frac{4\pm\sqrt[]{4\cdot14}}{-4} \\ \text{Separating the roots:} \\ x=\frac{4\pm2\sqrt[]{14}}{-4} \\ \end{gathered}[/tex]Dividing by -2:
[tex]x=\frac{-2\pm\sqrt[]{14}}{2}[/tex]Separating the solutions:
[tex]\begin{gathered} x_1=\frac{-2+\sqrt[]{14}}{2} \\ x_2=\frac{-2-\sqrt[]{14}}{2} \end{gathered}[/tex]