If it's not what You are looking for type in the equation solver your own equation and let us solve it.
2x^2=3x+45
We move all terms to the left:
2x^2-(3x+45)=0
We get rid of parentheses
2x^2-3x-45=0
a = 2; b = -3; c = -45;
Δ = b2-4ac
Δ = -32-4·2·(-45)
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{41}}{2*2}=\frac{3-3\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{41}}{2*2}=\frac{3+3\sqrt{41}}{4} $
| 3^a=69 | | x-0.35x=195 | | N=12.13-n | | 8-2x=5(x+4) | | f– 13= 13 | | -11/12=-5/6+c | | 8y=-2-40 | | y=–12–1 | | 3( | | 3( | | 3( | | 3(9x+7)= | | 3x–18=–9x+30 | | -2.1=r-8.5 | | 7c−–11=74 | | 7p+12=19 | | h(-4)=4-4 | | 5/k=5 | | {r}{10}=5 | | (2y-8)^2=18 | | g(-5)=3.4(-5)-8 | | h(-5)=4-4 | | 2x40=2x4 | | |-3x+4|=28 | | f(5)=4+1 | | f(2)=4+1 | | 9x-4x-3=4x+3+12 | | f(1)=4+1 | | 5x+25+6x+5+130-3x=180 | | 18g-14=12-20g | | 4+x+72=180 | | (4x-1)^2-16x+4=0 |