M(x)=-4/5x+-2;M=15

Solution for M(x)=-4/5x+-2;M=15 equation:

(M)=-4/5M+-2=15

We move all terms to the left:

(M)-(-4/5M+-2)=0


Domain of the equation: 5M+-2)!=0

M∈R


We add all the numbers together, and all the variables

M-(-4/5M-2)=0

We get rid of parentheses

M+4/5M+2=0

We multiply all the terms by the denominator


M*5M+2*5M+4=0

Wy multiply elements

5M^2+10M+4=0

a = 5; b = 10; c = +4;
Δ = b2-4ac
Δ = 102-4·5·4
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*5}=\frac{-10-2\sqrt{5}}{10}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*5}=\frac{-10+2\sqrt{5}}{10}
$