F(t)=(t+4)(t-2.3)

Solution for F(t)=(t+4)(t-2.3) equation:

(F)=(F+4)(F-2.3)

We move all terms to the left:

(F)-((F+4)(F-2.3))=0

We multiply parentheses ..

-((+F^2-2.3F+4F-9.2))+F=0


We calculate terms in parentheses: -((+F^2-2.3F+4F-9.2)), so:

(+F^2-2.3F+4F-9.2)

We get rid of parentheses

F^2-2.3F+4F-9.2

We add all the numbers together, and all the variables

F^2+1.7F-9.2

Back to the equation:

-(F^2+1.7F-9.2)


We add all the numbers together, and all the variables

F-(F^2+1.7F-9.2)=0

We get rid of parentheses

-F^2+F-1.7F+9.2=0

We add all the numbers together, and all the variables

-1F^2-0.7F+9.2=0

a = -1; b = -0.7; c = +9.2;
Δ = b2-4ac
Δ = -0.72-4·(-1)·9.2
Δ = 37.29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.7)-\sqrt{37.29}}{2*-1}=\frac{0.7-\sqrt{37.29}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.7)+\sqrt{37.29}}{2*-1}=\frac{0.7+\sqrt{37.29}}{-2}
$